Gravitation
NCERT Solutions • Class 9 Science • Chapter 9Chapter Exercises
1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Inverse Square Law
According to the Universal Law of Gravitation, Force $F \propto \frac{1}{d^2}$.
If distance $d$ becomes $d/2$, the new force $F’$ will be:
$F’ \propto \frac{1}{(d/2)^2} = \frac{1}{d^2/4} = 4 \times \frac{1}{d^2}$.
The force becomes 4 times the original value.
According to the Universal Law of Gravitation, Force $F \propto \frac{1}{d^2}$.
If distance $d$ becomes $d/2$, the new force $F’$ will be:
$F’ \propto \frac{1}{(d/2)^2} = \frac{1}{d^2/4} = 4 \times \frac{1}{d^2}$.
The force becomes 4 times the original value.
2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
The acceleration due to gravity ($g$) is given by $g = \frac{GM}{R^2}$, where $M$ is the mass of the Earth. This formula does not depend on the mass of the falling object ($m$).
Therefore, all objects, regardless of their mass, fall with the same acceleration (in the absence of air resistance).
Therefore, all objects, regardless of their mass, fall with the same acceleration (in the absence of air resistance).
3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of earth is $6 \times 10^{24}$ kg and radius is $6.4 \times 10^6$ m)
Given: $M = 6 \times 10^{24}$ kg, $m = 1$ kg, $R = 6.4 \times 10^6$ m, $G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$.
Formula: $F = \frac{GMm}{R^2}$
$F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2}$
$F \approx$ $9.8 \text{ N}$.
Formula: $F = \frac{GMm}{R^2}$
$F = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2}$
$F \approx$ $9.8 \text{ N}$.
4. The earth and the moon are attracted to each other… Does the earth attract the moon with a force that is greater or smaller or the same…? Why?
The Same Force.
According to Newton’s Third Law of Motion, action and reaction forces are equal and opposite. The gravitational force is a mutual interaction pair; the force with which Earth attracts the Moon is equal in magnitude to the force with which the Moon attracts the Earth.
According to Newton’s Third Law of Motion, action and reaction forces are equal and opposite. The gravitational force is a mutual interaction pair; the force with which Earth attracts the Moon is equal in magnitude to the force with which the Moon attracts the Earth.
5. If the moon attracts the earth, why does the earth not move towards the moon?
Inertia & Acceleration
Though the force is equal, the acceleration produced depends on mass ($a = F/m$). The mass of the Earth is huge compared to the Moon. Therefore, the acceleration produced in the Earth is negligible and not noticeable.
Though the force is equal, the acceleration produced depends on mass ($a = F/m$). The mass of the Earth is huge compared to the Moon. Therefore, the acceleration produced in the Earth is negligible and not noticeable.
6. What happens to the force between two objects, if:
(i) Mass of one object doubled: Force becomes 2 times ($F \propto M$).
(ii) Distance doubled: Force becomes 1/4th ($F \propto 1/2^2$).
Distance tripled: Force becomes 1/9th ($F \propto 1/3^2$).
(iii) Masses of both doubled: Force becomes 4 times ($F \propto 2M \times 2m$).
(ii) Distance doubled: Force becomes 1/4th ($F \propto 1/2^2$).
Distance tripled: Force becomes 1/9th ($F \propto 1/3^2$).
(iii) Masses of both doubled: Force becomes 4 times ($F \propto 2M \times 2m$).
Definitions & Concepts (Q7 – Q9)
7. Importance of Universal Law: It explains phenomena like the binding of us to earth, motion of moon around earth, motion of planets around sun, and tides.
8. Acceleration of Free Fall: The acceleration produced in a freely falling body due to the gravitational force of the earth. Value is approx $9.8 \text{ m/s}^2$.
9. Gravitational Force Name: The force between Earth and an object is called Gravity or Weight.
8. Acceleration of Free Fall: The acceleration produced in a freely falling body due to the gravitational force of the earth. Value is approx $9.8 \text{ m/s}^2$.
9. Gravitational Force Name: The force between Earth and an object is called Gravity or Weight.
10. Amit buys few grams of gold at the poles… Will the friend agree with the weight of gold bought at the equator? If not, why?
No, the friend will not agree.
The weight of an object ($W = mg$) depends on the value of $g$. The value of $g$ is greater at the poles ($g_p$) than at the equator ($g_e$) because the Earth’s radius is smaller at the poles. Therefore, the gold will weigh less at the equator than at the poles.
The weight of an object ($W = mg$) depends on the value of $g$. The value of $g$ is greater at the poles ($g_p$) than at the equator ($g_e$) because the Earth’s radius is smaller at the poles. Therefore, the gold will weigh less at the equator than at the poles.
11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
This is due to Air Resistance. The flat sheet of paper has a larger surface area, experiencing more air resistance, which slows it down. The crumpled ball has a smaller surface area, experiences less air resistance, and falls faster.
12. Gravitational force on the moon is 1/6th of earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?
On Earth:
$W_e = m \times g = 10 \times 9.8 =$ $98 \text{ N}$.
On Moon:
$W_m = \frac{1}{6} \times W_e = \frac{98}{6} \approx$ $16.33 \text{ N}$.
$W_e = m \times g = 10 \times 9.8 =$ $98 \text{ N}$.
On Moon:
$W_m = \frac{1}{6} \times W_e = \frac{98}{6} \approx$ $16.33 \text{ N}$.
13. A ball is thrown vertically upwards with a velocity of $49 \text{ m/s}$. Calculate (i) max height, (ii) total time.
Given: $u = 49 \text{ m/s}, v = 0, g = -9.8 \text{ m/s}^2$.
(i) Max Height ($h$)
$v^2 – u^2 = 2gh \Rightarrow 0 – (49)^2 = 2(-9.8)h$
$h = \frac{49 \times 49}{19.6} =$ $122.5 \text{ m}$.
(ii) Total Time
Time to go up ($t$): $v = u + gt \Rightarrow 0 = 49 – 9.8t \Rightarrow t = 5 \text{ s}$.
Total time = Time up + Time down = $5 + 5 =$ $10 \text{ s}$.
(i) Max Height ($h$)
$v^2 – u^2 = 2gh \Rightarrow 0 – (49)^2 = 2(-9.8)h$
$h = \frac{49 \times 49}{19.6} =$ $122.5 \text{ m}$.
(ii) Total Time
Time to go up ($t$): $v = u + gt \Rightarrow 0 = 49 – 9.8t \Rightarrow t = 5 \text{ s}$.
Total time = Time up + Time down = $5 + 5 =$ $10 \text{ s}$.
14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Given: $u = 0, s = 19.6 \text{ m}, g = 9.8 \text{ m/s}^2$.
Formula: $v^2 – u^2 = 2gs$
$v^2 – 0 = 2 \times 9.8 \times 19.6$
$v^2 = 19.6 \times 19.6 \Rightarrow v =$ $19.6 \text{ m/s}$.
Formula: $v^2 – u^2 = 2gs$
$v^2 – 0 = 2 \times 9.8 \times 19.6$
$v^2 = 19.6 \times 19.6 \Rightarrow v =$ $19.6 \text{ m/s}$.
15. A stone is thrown vertically upward with initial velocity $40 \text{ m/s}$. ($g = 10 \text{ m/s}^2$). Find max height, net displacement, total distance.
Given: $u = 40, v = 0, g = -10$.
Max Height ($h$)
$v^2 – u^2 = 2gh \Rightarrow -1600 = 2(-10)h \Rightarrow h = 80 \text{ m}$.
Net Displacement
Stone returns to starting point. Displacement = 0.
Total Distance
Up + Down = $80 + 80 =$ $160 \text{ m}$.
Max Height ($h$)
$v^2 – u^2 = 2gh \Rightarrow -1600 = 2(-10)h \Rightarrow h = 80 \text{ m}$.
Net Displacement
Stone returns to starting point. Displacement = 0.
Total Distance
Up + Down = $80 + 80 =$ $160 \text{ m}$.
16. Calculate the force of gravitation between the earth and the Sun.
Given: $M_E = 6 \times 10^{24} \text{ kg}, M_S = 2 \times 10^{30} \text{ kg}, d = 1.5 \times 10^{11} \text{ m}$.
$F = \frac{G M_E M_S}{d^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^2}$
$F = \frac{80.04 \times 10^{43}}{2.25 \times 10^{22}} \approx$ $3.56 \times 10^{22} \text{ N}$.
$F = \frac{G M_E M_S}{d^2} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^{30}}{(1.5 \times 10^{11})^2}$
$F = \frac{80.04 \times 10^{43}}{2.25 \times 10^{22}} \approx$ $3.56 \times 10^{22} \text{ N}$.
17. A stone is allowed to fall from a tower 100 m high and at the same time another stone is projected vertically upwards… with velocity 25 m/s. Calculate when and where the two stones will meet.
Let them meet at time $t$.
Falling Stone: Distance $x = \frac{1}{2}gt^2$.
Rising Stone: Distance $(100-x) = 25t – \frac{1}{2}gt^2$.
Adding both equations: $100 = 25t \Rightarrow$ $t = 4 \text{ s}$.
Position: $x = \frac{1}{2} \times 9.8 \times (4)^2 = 4.9 \times 16 = 78.4 \text{ m}$.
They meet 78.4 m from the top (or 21.6 m from the ground).
Falling Stone: Distance $x = \frac{1}{2}gt^2$.
Rising Stone: Distance $(100-x) = 25t – \frac{1}{2}gt^2$.
Adding both equations: $100 = 25t \Rightarrow$ $t = 4 \text{ s}$.
Position: $x = \frac{1}{2} \times 9.8 \times (4)^2 = 4.9 \times 16 = 78.4 \text{ m}$.
They meet 78.4 m from the top (or 21.6 m from the ground).
18. A ball thrown up vertically returns to the thrower after 6 s. Find (a) velocity, (b) max height, (c) position after 4 s.
Time to go up = $6/2 = 3 \text{ s}$.
(a) Initial Velocity ($u$) $v=u+gt \Rightarrow 0=u-9.8(3) \Rightarrow u=$ $29.4 \text{ m/s}$.
(b) Max Height ($h$) $h = ut + \frac{1}{2}gt^2 \Rightarrow$ using downward fall for simplicity: $h = \frac{1}{2}(9.8)(3)^2 =$ $44.1 \text{ m}$.
(c) Position after 4 s
At 4s, the ball has been falling for $4 – 3 = 1 \text{ s}$ from the top.
Distance fallen $= \frac{1}{2}(9.8)(1)^2 = 4.9 \text{ m}$.
Height from ground $= 44.1 – 4.9 =$ $39.2 \text{ m}$.
(a) Initial Velocity ($u$) $v=u+gt \Rightarrow 0=u-9.8(3) \Rightarrow u=$ $29.4 \text{ m/s}$.
(b) Max Height ($h$) $h = ut + \frac{1}{2}gt^2 \Rightarrow$ using downward fall for simplicity: $h = \frac{1}{2}(9.8)(3)^2 =$ $44.1 \text{ m}$.
(c) Position after 4 s
At 4s, the ball has been falling for $4 – 3 = 1 \text{ s}$ from the top.
Distance fallen $= \frac{1}{2}(9.8)(1)^2 = 4.9 \text{ m}$.
Height from ground $= 44.1 – 4.9 =$ $39.2 \text{ m}$.
19. In what direction does the buoyant force on an object immersed in a liquid act?
20. Why does a block of plastic released under water come up to the surface of water?
20. Why does a block of plastic released under water come up to the surface of water?
19. Direction: Vertically upwards.
20. Reason: The density of plastic is less than that of water. Therefore, the buoyant force (upthrust) acting on it is greater than its weight (gravity), pushing it up to the surface.
20. Reason: The density of plastic is less than that of water. Therefore, the buoyant force (upthrust) acting on it is greater than its weight (gravity), pushing it up to the surface.
21. Volume of 50 g substance is 20 cm³. Density of water 1 g cm⁻³. Float or sink?
22. Volume of 500 g packet is 350 cm³. Float or sink? Mass of water displaced?
22. Volume of 500 g packet is 350 cm³. Float or sink? Mass of water displaced?
21. Substance:
Density $= \frac{50}{20} = 2.5 \text{ g/cm}^3$.
Since $2.5 > 1$, it will sink.
22. Packet:
Density $= \frac{500}{350} \approx 1.42 \text{ g/cm}^3$.
Since $1.42 > 1$, it will sink.
Mass of water displaced = Volume of packet $\times$ Density of water
$= 350 \text{ cm}^3 \times 1 \text{ g/cm}^3 =$ $350 \text{ g}$.
Density $= \frac{50}{20} = 2.5 \text{ g/cm}^3$.
Since $2.5 > 1$, it will sink.
22. Packet:
Density $= \frac{500}{350} \approx 1.42 \text{ g/cm}^3$.
Since $1.42 > 1$, it will sink.
Mass of water displaced = Volume of packet $\times$ Density of water
$= 350 \text{ cm}^3 \times 1 \text{ g/cm}^3 =$ $350 \text{ g}$.