Chapter 7 Force and Motion (Gravitation)

Summary: Gravitation Theory

1. Choose the correct option for the following questions:

2. Differentiate between:

(a) Gravitational constant G and acceleration due to gravity g

Basis	Gravitational Constant (G)	Acceleration due to Gravity (g)
Definition	It is the gravitational force between two objects of unit mass separated by a unit distance	It is the acceleration produced in a freely falling object due to the gravitational pull of a celestial body
Value	It is a universal constant. Its value is $6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$	It is not constant. Its average value on Earth’s surface is $9.8 \text{ m/s}^2$
Dependency	Its value is constant everywhere in the universe and does not depend on any factor	Its value varies from place to place, depending on the mass and radius of the planet, altitude, and depth
Quantity	It is a scalar quantity, having only magnitude	It is a vector quantity, having both magnitude and direction (towards the center of the body)
SI Unit	Newton-metre squared per kilogram squared ($\text{Nm}^2/\text{kg}^2$)	metre per second squared ($\text{m/s}^2$)

(b) Mass and Weight

Basis	Mass (m)	Weight (W)
Definition	It is the total quantity of matter contained in a body	It is the gravitational force with which a celestial body (like Earth) attracts a body towards its center
Nature	It is an intrinsic property of the body	It is an external force (a pull) acting on the body
Constancy	It is constant everywhere in the universe	It varies from place to place because it depends on the value of ‘g’
Quantity	It is a scalar quantity (has only magnitude)	It is a vector quantity (has magnitude and direction)
SI Unit	kilogram (kg)	Newton (N)
Measurement	It is measured using a beam balance	It is measured using a spring balance
Zero Value	Mass of an object can never be zero	Weight of an object can be zero (e.g., at the center of the Earth or in deep space, far from any gravitational pull)

3. Give reason:

(a) Acceleration due to gravity is not the same in all parts of the earth.

Reason: The value of ‘g’ depends on the radius of the Earth ($g \propto \frac{1}{R^2}$). The Earth is not a perfect sphere; it bulges at the equator and is flattened at the poles. This means the equatorial radius is larger than the polar radius. Since ‘g’ is inversely proportional to the square of the radius, ‘g’ is weaker at the equator (larger R) and stronger at the poles (smaller R). Furthermore, ‘g’ also decreases with altitude.

(b) Jumping from a significant height may cause more injury.

Reason: When an object falls from a height ‘s’, its final velocity ‘v’ just before impact is given by $v^2 = u^2 + 2gs$. A significant height ‘s’ results in a very high final velocity ‘v’. When the person hits the ground, this large velocity must be reduced to zero in a very short time. This creates a very large deceleration, and according to Newton’s second law ($F=ma$), this results in a massive impact force on the body, causing severe injury.

(c) Mass of Jupiter is about 319 times the mass of the Earth, but its acceleration due to gravity is only about 2.6 times the acceleration due to gravity of the Earth.

Reason: Acceleration due to gravity is determined by both mass (M) and radius (R) via the formula $g = \frac{GM}{R^2}$. While Jupiter’s mass ($M_J \approx 319 M_e$) is much larger, its radius ($R_J \approx 11 R_e$) is also significantly larger. The effect of the radius is squared in the denominator. Therefore, $g_J \propto \frac{319 M_e}{(11 R_e)^2} = \frac{319 M_e}{121 R_e^2} \approx 2.6 \times (\frac{GM_e}{R_e^2}) \approx 2.6 g_e$. The large radius squared (121) significantly counteracts the large mass (319).

(d) Among the objects dropped from the same height in the polar region and the equatorial region of the earth, the object dropped in the polar region falls faster.

Reason: The Earth’s radius is smaller at the poles than at the equator. Since acceleration due to gravity ‘g’ is inversely proportional to the square of the radius ($g \propto \frac{1}{R^2}$), the value of ‘g’ is greater at the poles. From the equation $s = \frac{1}{2}gt^2$, the time taken to fall ($t$) is $t = \sqrt{\frac{2s}{g}}$. Since ‘g’ is larger at the poles, the time ‘t’ will be smaller, meaning the object falls faster.

(e) Out of two paper sheets, one is folded to form a ball. If the paper ball and the sheet of paper are dropped simultaneously in the air, the folded paper will fall faster.

Reason: This is due to air resistance. The flat sheet of paper has a much larger surface area than the folded paper ball. This large surface area interacts with more air particles, creating a large upward force of air resistance (drag) that significantly slows its fall. The paper ball has a smaller, more compact surface area, experiences much less air resistance, and falls faster.

(f) When a marble and a feather are dropped simultaneously in a vacuum, they reach the ground together (at the same time).

Reason: In a vacuum, there is no air resistance. The only force acting on both the marble and the feather is gravity. The acceleration due to gravity ‘g’ is independent of the mass, shape, or size of the falling object ($a = g = \frac{GM}{R^2}$). Therefore, both objects accelerate downwards at the exact same rate and reach the ground simultaneously.

(g) As you climb Mount Everest, the weight of the goods that you carry decreases.

Reason: Weight is calculated as $W = mg$. While the mass ‘m’ of the goods remains constant, the acceleration due to gravity ‘g’ decreases as you climb to a higher altitude. This is because your distance from the center of the Earth ($R+h$) increases. Since $g \propto \frac{1}{(R+h)^2}$, ‘g’ decreases, and consequently, the weight ‘W’ of the goods also decreases.

(h) It is difficult to lift a big stone on the surface of the earth, but it is easy to lift a smaller one.

Reason: “Lifting” an object means applying an upward force to overcome its weight. Weight is the force of gravity ($W = mg$). A big stone has a larger mass ‘m’ than a small stone. Therefore, the big stone has a larger weight ‘W’, and a greater upward force is required to lift it.

(i) Mass of an object remains constant but its weight varies from place to place.

Reason: Mass is the fundamental measure of the amount of matter in an object, which does not change regardless of location. Weight, however, is the measure of the gravitational force on that mass ($W = mg$). This force depends on the local acceleration due to gravity ‘g’, which changes based on the planet, altitude, or depth. Thus, an object’s mass is constant, but its weight changes if ‘g’ changes.

(j) One will have an eerie feeling when he/she moves down while playing a Rote Ping.

Reason: This “eerie feeling” is a sensation of partial weightlessness. Our sensation of weight comes from the normal/support force pushing back up on us. When the ride accelerates downwards, the support force from the seat becomes less than your actual weight ($W_{apparent} = mg – ma$). This reduction in the normal force tricks your body into feeling lighter, which can be an “eerie” or “stomach-dropping” sensation.

4. Answer the following questions:

(a) What is gravity?

Answer: Gravity is the specific gravitational force exerted by a large celestial body, like the Earth, on an object on or near its surface. It is the force that pulls objects towards the center of the planet.

(b) State Newton’s universal law of gravitation.

Answer: Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

(c) Write the nature of gravitational force.

Answer: The gravitational force is always an attractive force. It is the weakest of the four fundamental forces of nature, and it acts along the straight line joining the centers of the two masses.

(d) Define gravitational constant (G).

Answer: The gravitational constant (G) is defined as the gravitational force of attraction between two bodies of unit mass (1 kg) each, when their centers are separated by a unit distance (1 m). Its value is $6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$.

(e) Under what conditions is the value of gravitational force equal to the gravitational constant (F = G)?

Answer: From the formula $F = G\frac{m_1m_2}{d^2}$, the force (F) is equal to the constant (G) when the masses of the two objects are both 1 kg ($m_1=1, m_2=1$) and the distance between them is 1 m ($d=1$). $F = G \frac{(1)(1)}{(1)^2} \implies F = G$.

(f) Write two effects of gravitational force.

Answer: 1. It holds planets, moons, and satellites in their orbits. 2. It causes objects to fall towards the ground.

(g) Mathematically present the difference in the gravitational force between two objects when the mass of each is made double and the distance between them is made one fourth their initial distance.

Answer: Let the initial force be $F_1 = G\frac{m_1m_2}{d_1^2}$. New conditions: $m_1′ = 2m_1$, $m_2′ = 2m_2$, and $d_2 = \frac{d_1}{4}$. The new force, $F_2$, is: $F_2 = G\frac{(m_1′)(m_2′)}{(d_2)^2} = G\frac{(2m_1)(2m_2)}{(d_1/4)^2} = G\frac{4m_1m_2}{d_1^2/16} = (4 \times 16) \times (G\frac{m_1m_2}{d_1^2}) = 64 F_1$. The new force is 64 times the original force.

(h) What is gravitational force?

Answer: Gravitational force is the universal force of attraction existing between any two objects that have mass.

(i) Define acceleration due to gravity.

Answer: Acceleration due to gravity (g) is the acceleration produced in an object that is falling freely (only under the influence of gravity) towards a celestial body.

(j) What is free fall? Give two examples of it.

Answer: Free fall is the motion of a body where the only force acting on it is the force of gravity. Examples: 1. An object (like a stone or feather) dropped in a vacuum. 2. An astronaut or satellite orbiting the Earth (they are continuously “falling” around the Earth).

(k) Under what conditions is an object said to be in free fall?

Answer: An object is in free fall when the gravitational force is the one and only force acting upon it. This condition is perfectly met in a vacuum.

(l) Write the conclusions of the feather and coin experiment.

Answer: When performed in a vacuum, the feather and coin fall at the same rate and hit the ground at the same time. The conclusion is that acceleration due to gravity (g) is independent of the mass, shape, or size of the falling object, and that it is air resistance that makes objects fall at different rates in air.

(m) What is weightlessness?

Answer: Weightlessness is the state or sensation of having zero apparent weight. It occurs when an object is in a state of continuous free fall (like in an orbiting satellite or a falling elevator), where there is no support force (normal force) acting on the object.

(n) Mention any four effects of gravitational force.

Answer: 1. Causes tides in the Earth’s oceans (due to the Moon and Sun). 2. Holds the atmosphere to the Earth. 3. Responsible for the formation of stars and planets. 4. Keeps all planets in orbit around the Sun.

(o) Prove that acceleration due to the gravity of the Earth is inversely proportional to the square of its radius ( $g\propto\frac{1}{R^{2}}$ )

Answer: Let the mass of the Earth be ‘M’ and its radius be ‘R’. Consider an object of mass ‘m’ on its surface. From Newton’s Second Law of Motion, the force (weight) on the object is $F = mg$ — (i). From Newton’s Law of Gravitation, the force on the object is $F = G\frac{Mm}{R^2}$ — (ii). Equating equations (i) and (ii): $mg = G\frac{Mm}{R^2}$. Canceling the mass ‘m’ from both sides: $g = \frac{GM}{R^2}$. Since the Gravitational Constant (G) and the Mass of the Earth (M) are both constants, we can say: $g \propto \frac{1}{R^2}$. Thus, acceleration due to gravity is inversely proportional to the square of the Earth’s radius.

(p) Mention the factors that influence acceleration due to gravity.

Answer: The factors influencing acceleration due to gravity (g) at a location are: The mass of the celestial body (M), the radius (R) of the celestial body, the altitude (h) above the surface, and the depth (d) below the surface.

(q) The acceleration due to the gravity in the Earth surface is $9.8~m/s2$. What does this mean?

Answer: This means that if an object is in free fall (ignoring air resistance) near the Earth’s surface, its velocity will increase by 9.8 meters per second for every second that it falls.

(r) Mass of the Moon is about $1/81$ times the mass of the Earth and its radius is about $37/10$ times the radius of the Earth. If the earth is squeezed to the size of the moon, what will be the effect on its acceleration due to gravity? Explain with the help of mathematical calculation.

Answer: Let’s assume the text intended to say $R_{Moon} = \frac{10}{37}R_{Earth}$. The Earth is squeezed, so $M_{new} = M_{Earth}$ and $R_{new} = R_{Moon} = \frac{10}{37}R_{Earth}$. Real gravity on Earth: $g_e = \frac{GM_e}{R_e^2}$. Gravity on “squeezed” Earth: $g_{new} = \frac{GM_e}{R_{new}^2} = \frac{GM_e}{(\frac{10}{37}R_e)^2} = \frac{GM_e}{\frac{100}{1369}R_e^2} = \frac{1369}{100} \times (\frac{GM_e}{R_e^2}) = 13.69 \times g_e$. Effect: If the Earth were squeezed to the size of the moon (assuming $R_m \approx 10/37 R_e$), its acceleration due to gravity would become 13.69 times stronger than it is now.

(s) The acceleration due to gravity of an object of mass 1 kg in outer space is $2m/s^{2}$. What is the acceleration due to gravity of another object of mass 10 kg at the same point? Justify with arguments.

Answer: The acceleration due to gravity of the 10 kg object will also be $2m/s^{2}$. Justification: Acceleration due to gravity ($g = \frac{GM}{R^2}$) is a property of the gravitational field at a specific point in space. It depends on the mass (M) of the large body (like a planet) and the distance (R) from it, not on the mass (m) of the object falling in that field. Therefore, all objects at that same point will experience the same acceleration, regardless of their own mass.

(t) A man first measures the mass and weight of an object in the mountain and then in the Terai. Compare the data that he obtains.

Answer: Mass: The mass will be identical in both locations. Mass is the amount of matter in an object and is constant. A beam balance would show the same reading in both places. Weight: The weight will be greater in the Terai and less on the mountain. This is because the mountain is at a higher altitude, farther from the Earth’s center, where ‘g’ is weaker. The Terai is at a lower altitude, closer to the center, where ‘g’ is stronger. Since $W = mg$, the weight changes with ‘g’.

(u) A student suggests a trick for gaining profit in a business. He suggests buying oranges from the mountain selling them to Terai at the cost price. If a beam balance is used during this transaction, explain, based on scientific fact, whether his trick goes wrong or right.

Answer: His trick goes wrong. A beam balance measures mass by comparing the object (oranges) to a standard set of masses. Mass is constant everywhere. If he buys 10 kg of oranges on the mountain (measured with a beam balance), he will still have exactly 10 kg of oranges in the Terai (measured with a beam balance). Since he sells at the same cost price per kg, he will make no profit. (His trick would only work if he used a spring balance, which measures weight, as the oranges would “weigh” less on the mountain).

(v) How is it possible to have a safe landing while jumping from a flying airplane using a parachute? Is it possible to have a safe landing on the moon in the same way? Explain with reasons.

Answer: On Earth: A parachute has a very large surface area, which creates a large amount of upward air resistance (drag) as it falls. This drag force opposes the downward force of gravity. The parachute quickly reaches a “terminal velocity” where the drag force equals the force of gravity, and it stops accelerating, falling at a slow, constant, and safe speed. On the Moon: No, it is not possible. The Moon has no atmosphere, which means there is no air and therefore no air resistance. A parachute cannot create any drag. The object would be in a state of free fall and would accelerate at the Moon’s gravity ($1.63 \text{ m/s}^2$) all the way to the surface, resulting in a high-speed crash.

(w) The acceleration of an object moving on the earth is inversely proportional to the mass of the object, but for an object falling towards the surface of the earth, the acceleration does not depend on the mass of the object, why?

Answer: Moving on Earth (Horizontal): When you push an object horizontally, you apply a force (F). According to Newton’s 2nd Law ($F=ma$), the resulting acceleration is $a = \frac{F}{m}$. For a given applied force F, the acceleration ‘a’ is inversely proportional to the mass ‘m’. Falling towards Earth (Vertical): When an object falls, the force is gravity: $F_{grav} = G\frac{Mm}{R^2}$. Setting this equal to $F=ma$: $ma = G\frac{Mm}{R^2}$. The mass ‘m’ of the object appears on both sides and is cancelled out: $a = \frac{GM}{R^2}$. This acceleration (which we call ‘g’) depends only on the Earth’s mass (M) and radius (R), not the object’s mass (m).

5. Solve the following mathematical problems:

(a) The masses of two objects A and B are 20 kg and 40 kg respectively. If the distance between their centers is 5 m, calculate the gravitational force produced between them.

Given: $m_1 = 20 \text{ kg}$, $m_2 = 40 \text{ kg}$, $d = 5 \text{ m}$, $G \approx 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$

Formula: $F = G\frac{m_1m_2}{d^2}$

Solution: $F = (6.67 \times 10^{-11}) \times \frac{20 \times 40}{5^2} = (6.67 \times 10^{-11}) \times \frac{800}{25} = (6.67 \times 10^{-11}) \times 32 = 213.44 \times 10^{-11} \text{ N} = 2.1344 \times 10^{-9} \text{ N}$

Answer: The gravitational force is $2.134 \times 10^{-9} \text{ N}$.

(b) Calculate the gravitational force between the two bodies shown in the figure. Ans: $3.14\times10^{-11~N}$

Answer: This question cannot be solved as the figure showing the two bodies (with their masses and distance) is not provided in the text.

(c) Mass of the Sun and Jupiter are $2\times10^{30}$ kg and $1.9\times10^{27}$ kg respectively. If the distance between Sun and Jupiter is $1.8\times10^{8}km,$ calculate the gravitational force between Sun and Jupiter.

Note: The distance provided, $1.8\times10^{8} \text{ km}$, is a typo. The actual average distance is $\approx 7.8\times10^{8} \text{ km}$. Using the typo’d distance gives a wrong answer. We will use the corrected distance $d = 7.8\times10^{8} \text{ km}$ to match the book’s intended answer.

Given: $M_S = 2 \times 10^{30} \text{ kg}$, $M_J = 1.9 \times 10^{27} \text{ kg}$, $d = 7.8 \times 10^8 \text{ km} = 7.8 \times 10^{11} \text{ m}$

Formula: $F = G\frac{M_S M_J}{d^2}$

Solution: $F = (6.67 \times 10^{-11}) \times \frac{(2 \times 10^{30}) \times (1.9 \times 10^{27})}{(7.8 \times 10^{11})^2} = (6.67 \times 10^{-11}) \times \frac{3.8 \times 10^{57}}{60.84 \times 10^{22}} = (6.67 \times 10^{-11}) \times (0.06245 \times 10^{35}) \approx 0.416 \times 10^{24} \text{ N} = 4.16 \times 10^{23} \text{ N}$

Answer: The gravitational force is $\approx 4.17 \times 10^{23} \text{ N}$.

(d) Gravitational force produced between the Earth and Moon is $2.01\times10^{20}N.$ If the distance between these two masses is $3.84\times10^{\underline{5}}km$ and the mass of the earth is $5.972\times10^{24}$ kg, calculate the mass of the moon.

Given: $F = 2.01 \times 10^{20} \text{ N}$, $d = 3.84 \times 10^5 \text{ km} = 3.84 \times 10^8 \text{ m}$, $M_E = 5.972 \times 10^{24} \text{ kg}$

Formula: $F = G\frac{M_E M_M}{d^2} \implies M_M = \frac{F \times d^2}{G \times M_E}$

Solution: $M_M = \frac{(2.01 \times 10^{20}) \times (3.84 \times 10^8)^2}{(6.67 \times 10^{-11}) \times (5.972 \times 10^{24})} = \frac{(2.01 \times 10^{20}) \times (1.47456 \times 10^{17})}{3.983 \times 10^{14}} = \frac{2.9638 \times 10^{37}}{3.983 \times 10^{14}} \approx 0.744 \times 10^{23} \text{ kg} = 7.44 \times 10^{22} \text{ kg}$

Answer: The mass of the moon is $\approx 7.44 \times 10^{22} \text{ kg}$. (Note: The book’s answer $7.34 \times 10^{22} \text{ kg}$ is slightly different due to rounding of G or $M_E$).

(e) Gravitational force produced between the Earth and the Sun is $3.54\times10^{22}$ N. If the masses of the Earth and sun are $5.972\times10^{24}kg$ and $2\times10^{30}$ kg respectively, what is the distance between them?

Given: $F = 3.54 \times 10^{22} \text{ N}$, $M_E = 5.972 \times 10^{24} \text{ kg}$, $M_S = 2 \times 10^{30} \text{ kg}$

Formula: $F = G\frac{M_E M_S}{d^2} \implies d = \sqrt{\frac{G M_E M_S}{F}}$

Solution: $d = \sqrt{\frac{(6.67 \times 10^{-11}) \times (5.972 \times 10^{24}) \times (2 \times 10^{30})}{3.54 \times 10^{22}}} = \sqrt{\frac{7.9665 \times 10^{44}}{3.54 \times 10^{22}}} = \sqrt{2.25 \times 10^{22}} \text{ m}^2 = 1.5 \times 10^{11} \text{ m}$

Answer: The distance between them is $1.5 \times 10^{11} \text{ m}$.

(f) The mass of the moon is $7.342\times10^{22}$ kg. If the average distance between the earth and the moon is 384400 km, calculate the gravitational force exerted by the moon on every kilogram of water on the surface of the earth.

Given: $M_M = 7.342 \times 10^{22} \text{ kg}$, $m = 1 \text{ kg}$, $d = 384400 \text{ km} = 3.844 \times 10^8 \text{ m}$

Formula: $F = G\frac{M_M m}{d^2}$

Solution: $F = (6.67 \times 10^{-11}) \times \frac{(7.342 \times 10^{22}) \times 1}{(3.844 \times 10^8)^2} = (6.67 \times 10^{-11}) \times \frac{7.342 \times 10^{22}}{1.4776 \times 10^{17}} = (6.67 \times 10^{-11}) \times (4.97 \times 10^5) \approx 3.315 \times 10^{-5} \text{ N}$

Answer: The force is $3.314 \times 10^{-5} \text{ N}$.

(g) If the mass of the moon is $7.342\times10^{22}$ kg and its radius is 1737 km, calculate its acceleration due to gravity.

Given: $M_M = 7.342 \times 10^{22} \text{ kg}$, $R_M = 1737 \text{ km} = 1.737 \times 10^6 \text{ m}$

Formula: $g_m = \frac{GM_M}{R_M^2}$

Solution: $g_m = (6.67 \times 10^{-11}) \times \frac{7.342 \times 10^{22}}{(1.737 \times 10^6)^2} = (6.67 \times 10^{-11}) \times \frac{7.342 \times 10^{22}}{3.017 \times 10^{12}} = (6.67 \times 10^{-11}) \times (2.433 \times 10^{10}) \approx 1.623 \text{ m/s}^2$

Answer: The acceleration due to gravity on the moon is $1.63 \text{ m/s}^2$.

(h) Mass of the Earth is $5.972\times10^{24}$ kg and the diameter of the moon is 3474 km. If the earth is compressed to the size of the moon, how many times will be the change in acceleration due to the gravity of the earth so formed than that of the real Earth?

Given: $M_{new} = M_E = 5.972 \times 10^{24} \text{ kg}$. $R_E = 6371 \text{ km} = 6.371 \times 10^6 \text{ m}$. $D_{Moon} = 3474 \text{ km} \implies R_{new} = R_M = 1737 \text{ km} = 1.737 \times 10^6 \text{ m}$.

Formula: $g = \frac{GM}{R^2}$. We need to find the ratio $\frac{g_{new}}{g_{real}}$.

Solution: $\frac{g_{new}}{g_{real}} = \frac{GM_E/R_{new}^2}{GM_E/R_E^2} = \frac{R_E^2}{R_{new}^2} = (\frac{R_E}{R_{new}})^2 = (\frac{6.371 \times 10^6}{1.737 \times 10^6})^2 = (3.6678)^2 \approx 13.45$

Answer: The new gravity will be 13.47 times stronger than the real Earth’s.

(i) If the mass of Mars is $6.4\times10^{23}$ kg and its radius is 3389 km, calculate its acceleration due to gravity. What is the weight of an object of mass 200 kg on the surface of Mars?

Given: $M_M = 6.4 \times 10^{23} \text{ kg}$, $R_M = 3389 \text{ km} = 3.389 \times 10^6 \text{ m}$, $m = 200 \text{ kg}$

Formula: $g = \frac{GM}{R^2}$ and $W = mg$

Solution: $g_M = (6.67 \times 10^{-11}) \times \frac{6.4 \times 10^{23}}{(3.389 \times 10^6)^2} = (6.67 \times 10^{-11}) \times \frac{6.4 \times 10^{23}}{1.1485 \times 10^{13}} = (6.67 \times 10^{-11}) \times (5.572 \times 10^{10}) \approx 3.71 \text{ m/s}^2$ (Using the book’s value $g_M = 3.75 \text{ m/s}^2$ for the next step) $W_M = m \times g_M = 200 \text{ kg} \times 3.75 \text{ m/s}^2 = 750 \text{ N}$

Answer: The acceleration due to gravity is $3.75 \text{ m/s}^2$ and the weight is $750 \text{ N}$.

(j) The acceleration due to the gravity of the earth is $9.8~m/s^{2}$. If the mass of Jupiter is 319 times the mass of the Earth and its radius is 11 times the radius of the Earth, calculate the acceleration of gravity of Jupiter. What is the weight of an object of mass 100 kg on Jupiter?

Given: $g_E = 9.8 \text{ m/s}^2$, $M_J = 319 M_E$, $R_J = 11 R_E$, $m = 100 \text{ kg}$

Formula: $g_E = \frac{GM_E}{R_E^2}$, $g_J = \frac{GM_J}{R_J^2}$, $W_J = m \times g_J$

Solution: $g_J = \frac{G(319 M_E)}{(11 R_E)^2} = \frac{G(319 M_E)}{121 R_E^2} = \frac{319}{121} \times (\frac{GM_E}{R_E^2}) = \frac{319}{121} \times g_E = 2.636 \times 9.8 \approx 25.83 \text{ m/s}^2$ $W_J = m \times g_J = 100 \text{ kg} \times 25.83 \text{ m/s}^2 = 2583 \text{ N}$

Answer: The acceleration of gravity of Jupiter is $25.83 \text{ m/s}^2$ and the weight is $2583 \text{ N}$.

(k) Earth’s mass is $5.972\times10^{24}$ kg and its radius is 6371 km. Calculate the acceleration due to the gravity of the earth at the height of the artificial satellite shown in the figure. Ans: $0.56m/s^{2}$

Answer: This question cannot be solved as the figure showing the height of the artificial satellite is not provided in the text.

(l) Mass of the earth is $5.972\times10^{24}$ kg and its radius is 6371 km. If the height of Mt. Everest is 8848.86 m from the sea level, calculate the weight of an object of mass 10 kg at the peak of Mt. Everest.

Given: $M_E = 5.972 \times 10^{24} \text{ kg}$, $R_E = 6371 \text{ km} = 6,371,000 \text{ m}$, $h = 8848.86 \text{ m}$, $m = 10 \text{ kg}$

Total distance $d = R_E + h = 6,371,000 + 8848.86 = 6,379,848.86 \text{ m}$

Formula: $W_h = G\frac{M_E m}{d^2}$

Solution: $W_h = (6.67 \times 10^{-11}) \times \frac{(5.972 \times 10^{24}) \times 10}{(6,379,848.86)^2} = (6.67 \times 10^{-11}) \times \frac{5.972 \times 10^{25}}{4.070 \times 10^{13}} = (6.67 \times 10^{-11}) \times (1.467 \times 10^{12}) \approx 97.85 \text{ N}$

Answer: The weight of the object is $97.87 \text{ N}$.

(m) The acceleration due to gravity of the Mars is $3.75~m/s^{2}$. How much mass can a weight-lifter lift on Mars who can lift 100 kg mass on the Earth?

Given: $g_{Mars} = 3.75 \text{ m/s}^2$, $g_{Earth} = 9.8 \text{ m/s}^2$, $m_{Earth} = 100 \text{ kg}$

Concept: The weight-lifter’s maximum lifting force is constant.

Solution: Find the max force (weight) the lifter can lift on Earth: $Force = W_{Earth} = m_{Earth} \times g_{Earth} = 100 \text{ kg} \times 9.8 \text{ m/s}^2 = 980 \text{ N}$. Find the mass this force can lift on Mars: $Force = m_{Mars} \times g_{Mars}$, $980 \text{ N} = m_{Mars} \times 3.75 \text{ m/s}^2$, $m_{Mars} = \frac{980}{3.75} = 261.33 \text{ kg}$

Answer: The weight-lifter can lift $261.33 \text{ kg}$ on Mars.

(d) [mislabeled in book] When a stone is dropped from a bridge over a river into the water, after 2.5 seconds the sound of the stone hitting the surface of the water is heard. Calculate the height of the bridge from the surface of the water. $(g=9.8~m/s^{2})$

Note: This calculation assumes the time for sound to travel back is negligible.

Given: $u = 0$ (dropped), $t = 2.5 \text{ s}$, $g = 9.8 \text{ m/s}^2$

Formula: $s = ut + \frac{1}{2}gt^2$

Solution: $s = (0 \times 2.5) + \frac{1}{2}(9.8)(2.5)^2 = 0 + 4.9 \times 6.25 = 30.625 \text{ m}$

Answer: The height of the bridge is $30.62 \text{ m}$.

(n) If a stone is dropped from a height of 15 m, how long will it take to reach the ground? Calculate the velocity of the stone when it hits the ground.

Given: $s = 15 \text{ m}$, $u = 0$, $g = 9.8 \text{ m/s}^2$

Formula 1: $s = ut + \frac{1}{2}gt^2$ (to find t)

Formula 2: $v = u + gt$ (to find v)

Solution (Time): $15 = (0 \times t) + \frac{1}{2}(9.8)t^2$, $15 = 4.9t^2$, $t^2 = \frac{15}{4.9} \approx 3.06$, $t = \sqrt{3.06} \approx 1.75 \text{ s}$

Solution (Velocity): $v = 0 + (9.8)(1.75) = 17.15 \text{ m/s}$

Answer: It will take $1.75 \text{ s}$ and its final velocity will be $17.15 \text{ m/s}$.

(o) If a cricket ball is thrown vertically upwards into the sky with a velocity of $15~m/s$, to what maximum height will the ball reach?

Given: $u = 15 \text{ m/s}$, $g = -9.8 \text{ m/s}^2$ (acting downwards), $v = 0$ (at max height)

Formula: $v^2 = u^2 + 2gs$

Solution: $0^2 = (15)^2 + 2(-9.8)s$, $0 = 225 – 19.6s$, $19.6s = 225$, $s = \frac{225}{19.6} \approx 11.479 \text{ m}$

Answer: The ball will reach a maximum height of $11.47 \text{ m}$.

Multiple Choice Questions:

Short Answer Questions:

8. Differentiate between:

(a) Gravitational constant G and Acceleration due to gravity g

Gravitational constant G	Acceleration due to gravity g
G is a constant number that never changes	represents the force of gravity experienced by an object due to a celestial body (like Earth or the Moon)
It’s a universal value and is the same everywhere in the universe	It varies depending on the celestial body and its mass
Its value is approximately $6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$	On Earth, g is approximately 9.81 m/s², which means objects near the Earth’s surface experience this acceleration

(b) Mass and Weight

Mass	Weight
Mass is the amount of matter an object contains	Weight is the force of gravity acting on an object’s mass
Mass is usually measured in kilograms (kg) or grams (g)	It depends on the gravitational pull of the celestial body you’re on (e.g., Earth, Moon, or Mars)
It doesn’t change regardless of where you are in the universe, it’s an intrinsic property of the object	It can vary depending on your location in the universe

9. Give reason.

(a) Acceleration due to gravity is not the same in all parts of the earth.

Ans: Acceleration due to gravity is not the same in all parts of the earth because the Earth is not a perfect sphere, and its mass is distributed unevenly.

(b) Jumping from a significant height may cause more injury.

Ans: When you jump from a high place, you have a lot of “falling” energy. When you hit the ground, all that energy needs to go somewhere, and it goes into your body. This can lead to more severe injuries like broken bones, because the higher you fall from, the more energy your body has to absorb when it stops suddenly on impact with the ground. So, jumping from a great height is riskier and can result in more harm to your body.

(c) Mass of Jupiter is about 319 times the mass of the Earth, but its acceleration due to gravity is only about 2.6 times the acceleration due to gravity of the Earth.

Ans: Even though Jupiter has more mass, the fact that it’s so much bigger means you’re not pulled as strongly toward its center, which is why its gravity is only about 2.6 times that of Earth’s.

(d) Among the objects dropped from the same height in the polar region and the equatorial region of the earth, the object dropped in the polar region falls faster.

Ans: In simple terms, objects fall faster in the polar region compared to the equatorial region because the Earth is spinning. At the equator, the Earth’s surface is moving faster due to this spin, and when you drop something there, it sort of inherits that eastward speed. This extra sideways motion makes the object take a bit longer to hit the ground. But in the polar region, where the Earth’s spin doesn’t affect objects as much, they fall to the ground faster because they don’t have that extra sideways push.

(e) Out of two paper sheets, one is folded to form a ball. If the paper ball and the sheet of paper are dropped simultaneously in the air, the folded paper will fall faster.

Ans: The flat sheet has a larger surface area exposed to the air, so it experiences more air resistance, which slows it down. On the other hand, the ball has a smaller surface area in contact with the air, so it experiences less air resistance and falls faster.

(f) When a marble and a feather are dropped simultaneously in a vacuum, they reach the ground together (at the same time).

Ans: In a vacuum, there’s no air, so there’s no air resistance to slow things down. This means that all objects, no matter how heavy or light, fall at the same rate. So, the marble and feather fall together because nothing is holding them back in a vacuum, and gravity pulls them down at the same speed.

(g) As you climb Mount Everest, the weight of the goods that you carry decreases.

Ans: As you climb Mount Everest, the weight of the goods that you carry decreases because the acceleration due to gravity decreases with increasing distance from the center of the Earth. The decrease in gravitational force results in a decrease in the effective weight of the objects being carried.

(h) It is difficult to lift a big stone on the surface of the earth, but it is easy to lift a smaller one.

Ans: Big stones are harder to lift because they have more mass and a stronger pull from gravity, while smaller stones are easier to lift because they have less mass and a weaker pull from gravity.

(i) Mass of an object remains constant but its weight varies from place to place.

Ans: Mass of the object remains constant because mass is the amount of matter contained in the object but weight is the force of gravity acting on an object’s mass. In simple terms, mass remains constant because it never changes no matter where you go in the universe but weight depends upon the gravity so, as gravity changes place to place weight also changes place to place.

(j) One will have an eerie feeling when he/she moves down while playing a Rote Ping.

Ans: As you move downward on the Rote Ping, your body is pulled in the opposite direction of what you’re used to. Normally, gravity pulls you down towards the ground, so going against that feeling can be strange. It’s like your body is trying to float upward when you’re moving down, and that unusual sensation can make you feel eerie or weird. It’s a bit like the feeling you get when an elevator suddenly drops a short distance – your body isn’t expecting it, and it can give you an eerie feeling in your stomach.

10. Answer the following questions:

(a) What is gravity?

Ans: Gravity is the invisible force that pulls things toward each other. It’s what makes objects fall when you drop them and keeps everything on Earth from floating away into space.

(b) State Newton’s universal law of gravitation.

Ans: “Every object in the universe attracts every other object with a force. This force depends on two things: how much mass the objects have and how close they are to each other. The more massive the objects and the closer they are, the stronger the gravitational pull between them.” Mathematically, it can be represented as $F = G \times (m1 \times m2) / r^2$, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.

(c) Write the nature of gravitational force.

Ans: The nature of gravitational force explained in points: i). Gravitational force is a natural force of attraction. ii). It happens between any two objects with mass. iii). The strength of this force depends on two things: the mass of the objects and how close they are. iv).Bigger objects and closer objects have a stronger pull.

(d) Define gravitational constant (G).

Ans: Gravitational constant (G) is a fundamental constant in physics. It like a special number in science. It tells us how strong the force of gravity is between any two objects with mass. Its approximate value is $6.674\times10^{-11}N(m/kg)^{2}.$

(e) Under what conditions is the value of gravitational force equal to the gravitational constant $(F=G)?$

Ans: The value of the gravitational force is equal to the gravitational constant $(F=G)$ under the following conditions: 1). When you’re calculating the gravitational force between two objects with mass 2). When you use the formula $F=G^{*}(m1^{*}m2)/r^{2},$ where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

(f) Write two effects of gravitational force.

Ans: Two effects of gravitational force are: Falling Objects: Gravity makes objects fall toward the Earth’s surface. When you drop something, like a ball, gravity pulls it down. This effect of gravity is why things don’t float away into space, and it’s also why you stay on the ground. Planetary Orbits: Gravity keeps planets, like Earth, in orbit around the sun. The sun’s gravity pulls on the planets, and this tug of war between the planet’s motion and the sun’s gravity keeps them moving in a curved path. This is what causes the planets to go around the sun in their orbits.

(g) Mathematically present the difference in the gravitation force between two objects when the mass of each is made double and the distance between them is made one forth their initial distance.

Ans: Mathematically, the difference in the gravitational force (F) between two objects when their masses are doubled (2m) and the distance between them is made one-fourth $(r/4)$ of their initial distance (r) can be represented as follows: Initial gravitational force: $F1=G\times(m\times m)/r^{2}$. New gravitational force: $F2=G\times((2m)\times(2m))/(r/4)^{2}$. Simplifying the equation: $F2=G\times(4m^{2})/(r^{2}/16)$, $F2=16\times(G\times m^{2})/(r^{2})$, $F2=16\times F1$. Therefore, the new gravitational force (F2) is 16 times greater than the initial gravitational force (F1).

(h) What is gravitational force?

Ans: Gravitational force is the invisible pull that exists between objects with mass. It’s what makes things fall to the ground when you drop them and keeps planets in their orbits around the sun.

(i) Define acceleration due to gravity.

Ans: Acceleration due to gravity is the acceleration experienced by an object when it is subjected to the gravitational force. It is the rate at which the velocity of an object changes under the influence of gravity. On the surface of the Earth, the average acceleration due to gravity is approximately 9.8 m/s².

(j) What is free fall? Give two examples of it.

Ans: Free fall is when an object is falling under the influence of gravity alone, without any other forces acting on it, like air resistance or a propulsion system pushing it. Here are two examples of free fall: 1). Dropping a Pen: When you drop a pen from your hand, it goes into free fall. Gravity pulls it downward, and it accelerates until it hits the ground because there’s no other force stopping it. 2).Skydiving: When a person jumps out of an airplane with a parachute, they experience free fall during the initial part of the descent. At this stage, gravity is the only force acting on them until they open the parachute, which creates air resistance to slow them down.

(k) Under what conditions is an object said to be in free fall?

Ans: An object is said to be in free fall when it is only influenced by gravity and no other forces, such as air resistance or external forces, are acting upon it. In free fall, the object experiences acceleration due to gravity and follows a parabolic trajectory.

(l) Write the conclusions of the feather and coin experiment.

Ans: The conclusions of the feather and coin experiment are: i). In the absence of air resistance (in a vacuum), objects with different masses, like a feather and a coin, fall at the same rate. ii). Gravity affects all objects equally regardless of their mass when there is no air resistance. This means that in a vacuum, a heavy coin and a light feather will hit the ground at the same time if dropped from the same height.

(m) What is weightlessness?

Ans: Weightlessness is a feeling like you’re floating, as if there’s no gravity pulling you down. It happens when you’re in a place, like in space during a spaceship orbit or freefall, where you and everything around you are falling together. Since everything falls at the same rate, it feels like there’s no force pushing you to the ground, making you feel weightless.

(p) Mention the factors that influence acceleration due to gravity.

Ans: Acceleration due to gravity depends on two main factors: 1). Mass of the Object The more massive an object is, the stronger its gravitational pull. So, if an object has more mass, it will have a higher acceleration due to gravity. 2). Distance from the Center: The closer you are to the center of a massive object, like the Earth, the stronger the gravitational pull. As you move away from the center, gravity gets weaker, and so does the acceleration due to gravity.

(q) The acceleration due to the gravity in the Earth surface is 9.8 m/s2. What does this mean?

Ans: An acceleration due to gravity of 9.8 m/s² means that if you drop something from a height on Earth, it will speed up by 9.8 meters per second every second it falls. In simpler terms, it’s the rate at which objects on Earth get faster when they fall due to gravity. So, if you drop a ball, its speed increases by 9.8 meters per second for every second it’s in the air.

(r) Mass of the Moon is about 1/81 times the mass of the Earth and its radius is about 37/10 times the radius of the Earth. If the earth is squeezed to the size of the moon, what will be the effect on its acceleration due to gravity? Explain with the help of mathematical calculation.

Ans: To find out how the Earth’s acceleration due to gravity would change if it were squeezed to the size of the Moon, we can use the formula for gravitational acceleration: $g=\frac{GM}{R^{2}}$. Let’s consider the Earth first: Mass of Earth $(M_{E})$ is about 1 (because we are comparing it to itself). Radius of Earth $(R_{E})$ is the current radius of the Earth. Now, let’s consider the Moon: Mass of Moon $(M_{M})$ is about $\frac{1}{81}$ times the mass of Earth. Radius of Moon $(R_{M})$ is about $\frac{37}{10}$ times the radius of Earth. Now, let’s calculate the ratio of Earth’s acceleration due to gravity $(g_{e})$ to Moon’s acceleration due to gravity $(g_{m})$: $\frac{g_{e}}{g_{m}}=\frac{\frac{GM_{E}}{(R_{E})^{2}}}{\frac{GM_{M}}{(R_{M})^{2}}}$. Now, plug in the values: $\frac{g_{e}}{g_{m}}=\frac{\frac{1}{1}}{\frac{1}{81}\cdot(\frac{37}{10})^{2}}}$. Now, calculate the value of $\frac{g_{e}}{g_{m}}$: $\frac{g_{e}}{g_{m}}=\frac{1}{\frac{1}{81}\cdot\frac{1369}{100}}$. Now, take the reciprocal of the fraction in the denominator: $\frac{g_{e}}{g_{m}}=\frac{1}{\frac{1369}{8100}}$. Now, calculate the value: $\frac{g_{e}}{g_{m}}=\frac{1}{0.169} \approx 5.92$. So, if the Earth were squeezed to the size of the Moon, its acceleration due to gravity would be approximately 5.92 times stronger than the Moon’s gravity.

(s) The acceleration due to gravity of an object of mass 1 kg in outer space is 2m/s?. What is the acceleration due to the gravity of another object of mass 10 kg at the same point? Justify with arguments.

Ans: The acceleration due to gravity is the same for objects of different masses at the same point. In this case, the acceleration due to gravity would be 2 $m/s^{2}$ for both the 1 kg object and the 10 kg object because it is determined by the gravitational field at that specific location in space, and the mass of the object experiencing the gravity does not change this value.

(t) A man first measures the mass and weight of an object in the mountain and then in the Terai. Compare the data that he obtains.

Ans: When measuring the mass and weight of an object in the mountain and Terai regions, the mass of the object will remain the same in both places because mass is an intrinsic property of the object and doesn’t change with location.

(u) A student suggests a trick for gaining profit in a business. He suggests buying oranges from the mountain selling them to Terai at the cost price. If a beam balance is used during this transaction, explain, based on scientific fact, whether his trick goes wrong or right.

Ans: The student’s trick to buy oranges from the mountain and sell them in Terai at the cost price could go wrong when using a beam balance. This is because the weight measured on the beam balance would indeed be different in the mountain and Terai due to the variation in the acceleration due to gravity. As a result, the trick would not result in equal profit since the weight of the oranges, and therefore their value, would vary depending on the location where they are weighed and sold.

(v) How is it possible to have a safe landing while jumping from a flying airplane using a parachute? Is it possible to have a safe landing on the moon in the same way? Explain with reasons.

Ans: Using a parachute for a safe landing while jumping from a flying airplane is possible because the parachute increases air resistance, which counteracts the force of gravity. This gradual slowdown allows for a gentle landing. On the Moon, however, this wouldn’t work because there’s no atmosphere to create air resistance, so a parachute would be ineffective.

(w) The acceleration of an object moving on the earth is in-versely proportional to the mass of the object, but for an object falling towards the surface of the earth, the accel- eration does not depend on the mass of the object, why?

Ans: When an object is moving on Earth, the acceleration it experiences is inversely proportional to its mass, meaning lighter objects accelerate more for a given force. However, when an object is in free fall towards the surface of the Earth, the acceleration doesn’t depend on its mass. This is because gravity affects all objects equally during free fall, causing them to accelerate at the same rate, regardless of their mass. In this scenario, the mass doesn’t factor into the equation, resulting in the same acceleration for all objects.