TS Inter 1st Year Physics Study Material Chapter 9 Gravitation

Telangana TSBIE TS Inter 1st Year Physics Study Material 9th Lesson Gravitation Textbook Questions and Answers.

TS Inter 1st Year Physics Study Material 9th Lesson Gravitation

Very Short Answer Type Questions

Question 1.
State the unit and dimension of universal gravitational constant (G).
Answer:
Units of G = N-m² / kg².
Dimensional formula = M^-1 L³ T^-2.

Question 2.
State the vector form of Newton’s law of gravitation.
Answer:
Vector form of Newton’s Law of gravitation is \(\overline{\mathrm{F}}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2 \overline{\mathrm{r}}}{\overline{\mathbf{r}}^3}\)

Question 3.
If the gravitational force of the Earth on the Moon is F. What is the gravitational force of the moon on the earth? Do these forces form an action-reaction pair?
Answer:
Gravitational force between earth and moon and moon and earth are same
i-e., F_EM = – F_ME

Gravitational force between the bodies are treated as action-reaction pair.

Question 4.
What would be the change in acceleration due to gravity (g) at the surface, if the radius of Earth decreases by 2% keeping the mass of Earth constant?
Answer:
Acceleration due to gravity, g = \(\frac{GM}{R^2}\)
When mass is kept as constant and radius
is decreased by 2% then \(\frac{\Delta \mathrm{R}}{\mathrm{R}}\) × 100 = 2

From distribution of errors in multiplications and divisions \(\frac{\Delta \mathrm{g}}{\mathrm{g}}\) × 100 = -2\(\frac{\Delta \mathrm{R}}{\mathrm{R}}\) × 100

% Change in g = – 2 × 2 = – 4% – ve sign indicates that when R decreases ‘g’ increases.

Question 5.
As we go from one planet to another, how will a) the mass and b) the weight of a body change?
Answer:

1. As we go from one planet to another planet mass of the body does not change. Mass of a body is always constant.
2. As we move from one planet to another planet weight of the body gradually decreases. It become weightless. When we approaches the other planet the weight will gradually increases.

Question 6.
Keeping the length of a simple pendulum constant, will the time period to be the same on all planets? Support your answer with reason.
Answer:
Even though length of pendulum / is same the time period of oscillation T value changes from planet to planet.
Time period of pendulum, T = 2π\(\sqrt{\frac{l}{g}}\)
i.e., T depends on l and g.

Acceleration due to gravity, (g = \(\frac{GM}{R^2}\)) changes from planet to planet.

Hence Time period of pendulum changes even though length ‘l’ is same.

Question 7.
Give the equation for the value of g at a depth ‘d’ from the surface of Earth. What is the value of ‘g’ at the centre of Earth?
Answer:
Acceleration due to gravity at a depth ‘d’ below the ground is, gd = g(1 – \(\frac{D}{R}\))
Acceleration due to gravity at centre of earth is zero. (Since D = R)

Question 8.
What are the factors that make ‘g’ the least at the equator and maximum at the poles?
Answer:
‘g’ is least at equator due to
1) The equatorial radius of earth is maximum ∵ g = \(\frac{GM}{R^2}\) (∵ R = maximum)

2) Due to rotation of earth centrifugal force will act on the bodies. It opposes gravitational pull of earth on the bodies. At equator centrifugal force is maximum. So ‘g’ value is least at equator.
The g’ ⇒ maximum at poles due to

1) The polar radius of earth is minimum
(∵ g = \(\frac{GM}{R^2}\))

2) Centrifugal force due to rotation of earth is zero at poles. This centrifugal force reduces earth’s gravitational pull.

Since Centrifugal force is zero, ‘g’ value is maximum at poles.

Question 9.
“Hydrogen is in abundance around the sun but not around earth”. Explain.
Answer:
The escape velocity on the sun is very high compared to that on the earth. The gravitational pull of the sun is very large because of its larger mass compared to that of the earth. So it is very difficult for hydrogen to escape from the Sun’s atmosphere. Hence hydrogen is abundant on sun.

Question 10.
What is the time period of revolution of a geostationary satellite? Does it rotate from West to East or from East to West?
Answer:
Time period of geostationary satellite is equal to time period of rotation of earth.

∴ Time period of geostationary orbit T = 24 hours. Satellites in geostationary orbit will revolve round the earth in west to east direction in an equatorial plane.

Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).

Question 11.
What are polar satellites?
Answer:
Polar satellite :
Polar satellites are low altitude satellites. They will revolve around the poles of the earth in a north-south direction. Time period of polar satellites is nearly 100 minutes.

Short Answer Questions

Question 1.
State Kepler’s Laws of planetary motion. [TS Mar. ’17]
Answer:
Kepler’s Laws :
Law of orbits (1st law):
All planets will move in elliptical orbits with the sun lies at one of its foci.

Law of areas (2nd law) :
The line that joins any planet to the sun sweeps equal areas in equal intervals of time, i.e., \(\frac{\Delta \mathrm{A}}{\Delta \mathrm{T}}\)= constant. i.e., planets will appear to move slowly when they are away from sun, and they will move fast when they are nearer to sun.

Law of periods (3rd law) :
The square of time period of revolution of a planet is proportional to the cube of the semi major axis of the ellipse traced out by the planet.
i.e., T² ∝ R³ or \(\frac{T^2}{R^3}\) = constant

Question 2.
Derive the relation between acceleration due to gravity (g) at the surface of a planet and Gravitational constant (G).
Answer:
Relation between g and G :
Each and everybody was attracted towards centre of earth with some force. This is called weight of the body,
W = mg …………. (1)

This force is due to gravitational pull on the body by the earth.

For small distances above earth from centre of earth is equal to radius of earth ‘R’.

According to Newton’s law of gravitation.

Question 3.
How does the acceleration due to gravity (g) change for the same values of height (h) and depth(d)?
Answer:
Variation of ‘g’ with altitude :
When we go to a height ‘h’ above the ground ‘g’ value decreases.
On surface of earth (g) = \(\frac{GM}{R^2}\)
At an altitude ‘h’ g(h) = \(\frac{GM}{(R+h)^{2}}\) because

R + h is the distance from centre of earth to the given point at ‘h’.
h << R ⇒ g(h) = g(1 – \(\frac{2h}{R}\))
So acceleration due to gravity decreases with height above the ground.

Variation of ‘g’ with depth :
When we go deep into the ground ‘g’ value decreases.

At a depth ‘d’ inside the ground mass of earth upto the point d from centre will exhibit force of attraction on the body. The remaining mass does not exhibit any influence. So effective radius is (R – d) only. Acceleration due to gravity ‘g’ at a depth ‘d’ is given by

So ‘g’ value decreases with depth below the ground.

Question 4.
What is orbital velocity? Obtain an expression for it. [AP Mar. 17, 14; May 18. 14]
Answer:
Orbital velocity (V₀) :
Velocity of a satellite moving in the orbit is called orbital velocityog.

Let a satellite of mass m is revolving round the earth in a circular orbit at a height ‘h’ above the ground.

Radius of the orbit = R + h where R is radius of earth.

In orbital motion is “The centrifugal and centripetal forces acting on the satellite”.

Centrifugal force = \(\frac{mV^2}{r}=\frac{mV^{2}_{0}}{R+h}\) ……… (1)
(In this case V = V₀ and r = R + h)

Centripetal force is the force acting towards the centre of the circle it is provided by gravitational force between the planet and satellite.

V₀ = \(\sqrt{gR}\) is called orbital velocity. Its value is 7.92 km/sec.

Question 5.
What is escape velocity? Obtain an expression for it. [TS Mar. ’19, ’16; AP Mar. ’19. ’18, ’15, ’13. May ’17, ’16]
Answer:
Escape speed (v₁)_min :
It is defined as the minimum velocity required by a body to overcome gravitational field of earth is known as escape velocity.

For a body of mass ‘m’ gravitational potential energy on surface of earth PE = – \(\frac{G.m.M_E}{R_E}\)

For a body to escape from gravitational field of earth its kinetic energy must be equal or more than gravitational potential energy.

Question 6.
What is a geostationary satellite? State [AP Mar. ’16, June ’15; May ’13; TS Mar. ’18, ’15, May ’18, ’16, June ’15]
Answer:
A geostationary satellite will always appears to be stationary relative to earth.

The time period of geostationary satellite is equals to time perfod of rotation of earth.

∴ Time period of geostationary orbit t = 24 hours. Satellites in geostationary orbit will revolve round the earth in west to east direction in an equatorial plane.

Uses of geostationary satellites:

1. For study of the upper layers of the atmosphere.
2. For forecasting the changes in atmosphere and weather.
3. For finding the size and shape of earth.
4. For investigating minerals and ores present in the earth’s crust.
5. For transmission of T.V. signals.
6. For study of transmission of radio waves.
7. For space research.

Question 7.
If two places are at the same height from the mean sea level; One is a mountain and other is in air. At which place will ‘g’ be greater? State the reason for your answer.
Answer:
‘g’ value on the mountain is greater than g’ value in air even though both are at same height.

For a point on mountain while deciding the ‘g’ value, mass of mountain is also considered which leads to change in ‘g’ value depending on local condition such as concentration of huge mass at a particular place. Whereas for a point in air no such effect is considered. Hence ‘g’ on the top of mountain is more.

Question 8.
The weight of an object is more at the poles than at the equator. At which of these can we get more sugar for the same weight? State the reason for your answer.
Answer:
If we are using common balance to measure sugar we will get some quantity of sugar both at equator and at poles.

Whereas if we are using spring balance to weigh sugar then weight of sugar at poles is more. So we will get less quantity.

Weight of sugar at equator is less. So we will get more quantity of sugar at equator.

Question 9.
If a nut becomes loose and gets detached from a satellite revolving around the earth, will it fall down to earth or will it revolve around earth? Give reasons for your answer.
Answer:
If a nut is detached from a satellite revolving in the orbit then its velocity is equals to orbital velocity. So it continues to revolve in the same orbit. It does not fall to earth.

Question 10.
An object projected with a velocity greater than or equal to 11.2 km.s^-1 will not return to earth. Explain the reason.
Answer:
Escape velocity of earth is 11.2 km/sec. If any body acquires a velocity of 11.2 km/ sec. or more its kinetic energy is more than gravitational potential energy. So earth is not able to stop the motion of that body. So any body with a velocity 11.2 km/s or more will escape from gravitational field of earth and never comes back to earth.

Long Answer Questions

Question 1.
Define gravitational potential energy and derive an expression for it associated with two particles of masses m₁ and m₂.
Answer:
Gravitational potential energy of a body at a point in a gravitational field of another. It is defined as the amount of work done in brining the given body from infinity to that point in the field is called Gravitational potential energy.

Expression for gravitational potential energy :
Consider two particles of masses m, and m2 are placed at the points O’ and p respectively. Let the distance between the two particles is r’ i.e., OP = r.

Let us calculate the gravitational potential energy of the particle of mass m₂ placed at point p in the gravitational field of m₁. Join OP and extended it in forward direction. Consider two points A and B on this line such that OA = x and AB = dx.

The gravitational force of attraction on the particle at A is, F = \(\frac{\mathrm{Gm_1m_2}}{\mathrm{x^2}}\)

Small amount of work done in bringing the particle without acceleration through a very small distance AB is, dW = F dx
= \(\frac{\mathrm{Gm_1m_2}}{\mathrm{x^2}}\)

Total workdone in bringing the particle from infinity to the point P is,

Since, this work done is stored in the particle as its gravitational potential energy (U). Therefore, gravitational potential energy of the particle of mass m₂ placed at point p’ in the gravitational field of particle of mass
m₁ at distance r is, U = \(\frac{\mathrm{-Gm_1m_2}}{\mathrm{r}}\)

Here, negative sign shows that the potential energy is due to attractive gravitational force between two particles.

Question 2.
Derive an expression for the variation of acceleration due to gravity (a) above and (h) below the surface of the Earth.
Answer:
Variation of acceleration due to gravity above the surface of earth :
We know ‘g’ on planet, g = \(\frac{GM}{R^2}\). But on earth g’ value changes with height above the ground h’.

Variation of ‘g’ with altitude :
For a point h’ above the earth total mass of earth seems to be concentrated at centre of earth. Now distance from centre of earth is (R + h).
Acceleration due to gravity at ‘h’ = g(h)
= \(\frac{GM_E}{(R_E+h)^2}\)

For small values of ‘h’ i.e., h << R than

Variation of acceleration due to gravity below the surface of earth :
At a depth d’ inside the ground mass of earth upto the point d from centre will exhibit force of attraction on the body. The remaining mass does not exhibit any influence.
Mass of spherical body M ∝ R³

∴ M_S/M_E = (R_E – d)³/ R³^E where M_s is mass of earth’s shell upto a depth ‘d’ from centre. Gravitational force at depth ‘d’ is

So ‘g’ value decreases with depth below the ground.

Question 3.
State Newton’s Universal Law of Gravitation. Explain how the value of the Gravitational constant (G) can be determined by Cavendish method.
Answer:
Newton’s law of gravitation :
Every body in the universe attracts every other body with a force which is directly proportional to theproduct of their masses and inversely proportional to the square of the distance between them.

This is always a force of attraction and acts along the line joining the two bodies.

Cavendish experiment to find gravitational constant ‘G’ :
Cavendish experiment consists of a long metalic rod AB to which two small lead spheres of mass ‘m’ are attached.

This rod is suspended from a rigid support with the help of a thin wire. Two heavy spheres of mass M are brought near to these small spheres in opposite direction. Then gravitational force will act between the spheres.

Force between the spheres, r = \(\frac{GMm}{d^2}\)

Two equal and opposite forces acting at the two ends of the rod AB will develop force couple and the rod will rotate through an angle ‘θ’.
∴ Torque on the rod F × L = \(\frac{GM.m}{d^2}\) → (1)
Restoring force couple = τθ → (2)
Where τ = Restoring couple per unit twist

By measuring 0 we can calculate ‘G’ value when other parameters are known. Practical value of G is 6.67 × 10^-11 Nm²/kg².

Problems

(Gravitational Constant ‘G’ = 6.67 × 10^-11 Nm²kg^-2; Radius of earth ‘R’ = 6400 km; Mass of earth ‘M_E‘ = 6 × 10²⁴ kg)

Question 1.
Two spherical balls each of mass 1 kg are placed 1 cm apart. Find the gravitational force of attraction between them.
Solution:
Mass of each ball, m = 1 kg;
Separation, r = 1 cm = 10^-2 m
Gravitational force of attraction,

Question 2.
The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of 10 cm, the gravitational force between them is 6.67 × 10^-7 N. Find the masses of the two balls.
Solution:
Mass of 1st ball = m;
Mass of 2nd ball = 4m.
Separation, r = 10 cm = 0.1 m;
Mass of the 1st ball = m = ?
Gravitational force, F = 6.67 × 10^-7 N
∴ Mass of the balls are 5 kg, 20 kg.

Question 3.
Three spherical balls of masses 1 kg, 2 kg and 3 kg are placed at the corners of an equilateral triangle of side 1 m. Find the magnitude of the gravitational force exerted by the 2 kg and 3 kg masses on the 1 kg mass.
Solution:
Side of equilateral triangle, a = lm.
Masses at corners = 1 kg, 2 kg, 3 kg.
Force between 1 kg, 2g = F₁ = G.\(\frac{2\times1}{1^2}\) = 2 G
Force between I kg, 3kg = F₂ = G.\(\frac{3\times1}{1^2}\)= 3G.
Now F₁ & F₂ act with an angle of 60°
∴ Resultant force

Question 4.
At a certain height above the earth’s surface, the acceleration due to gravity is 4% of its value at the surface of the earth. Determine the height.
Solution:
Acceleration due.to gravity at a height, h = 4% of g.
Radius of earth, R = 6400K.M. = 6.4 × 10⁶m.

Question 5.
A satellite orbits the earth at a height of 1000 km. Find its orbital speed.
Solution:
Radius of earth, R = 6,400 km = 6.4 × 10⁶ m;
Mass of earth, M = 6 × 10²⁴
Height of satellite h = 1000 km;
G = 6.67 × 10¹¹ N – m² /kg²

Question 6.
A satellite orbits the earth at a height equal to the radius of earth. Find it’s (i) orbital speed and (ii) Period of revolution.
Solution:
Radius of earth, R = 6400 k.m.;
-height above earth, h = R.
Mass of earth, M = 6 × 10²⁴
G = 6.67 × 10^-11 N – m²/kg²

Question 7.
The gravitational force of attraction between two objects decreases by 36% when the distance between them is increased by 4 m. Find the original distance between them.
Solution:
Let force between the objects = F;
Distance between them = r.
For Case II distance, r₁ = (r + 4);
New force, F₁ = 36% less than F

⇒ 100 r² = 64 (r + 4)² Take square roots on both sides.
10r = 8 (r + 4) ⇒ 10 r = 8r + 32
⇒ (10 – 8) r = 2r = 32
∴ r = 16 m.

Question 8.
Four identical masses m are kept at the corners of a square of side a. Find the gravitational force exerted on one of the masses by the other masses.
Solution:
Given all masses are equal
∴ m₁ = m₂ = m₃ = m₄
Force between m₁, m₄ = F₁ = \(\frac{G.m^2}{a^2}\) ……… (1)
Force between m₄, m₃ = F₂ = \(\frac{G.m^2}{a^2}\) ……… (2)
Forces F₁ and F₂ act perpendicularly.

Their magnitudes are equal.

Now forces F_R and F₃ are like parallel. So resultant is sum of these forces.
Total force at m₄ due to other masses

Question 9.
Two spherical balls of 1 kg and 4 kg are separated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.
Solution:
Mass, m₁ = 1 kg ; Mass, m₂ = 4 kg ;
Separation, d = 12 cm
Mass of 3rd body m₃ = ?

For m₃ not to experience any force the condition is
Force between m₁, m₃ = Force between m₂, m₃.

Take square roots on both sides,
d – x = 2x ⇒ d = 3x or x = \(\frac{12}{3}\) = 4 cm
∴ Distance from 1 kg mass = 4 cm

Question 10.
Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.
Solution:
Mass m and radius R are same for all spheres.
Force between 1, 3 spheres = F₁ = \(\frac{G.m^2}{(2R)^2}\)
Force between 1, 2 spheres = F₂ = \(\frac{G.m^2}{(2R)^2}\)

Now F₁ and F2 will act with an angle θ = 60° between them so from Parallelogram law

Question 11.
Two satellites are revolving round the earth at different heights. The ratio of their orbital speeds is 2 : 1. If one of them is at a height of 100 km what is the height of the other satellite?
Solution:
Mass of earth, m = 6 × 10²⁰ kg ;
G = 6.67 × 10^-11 N-m² / kg²
Ratio of orbital velocities V₀₁ : V₀₂ = 2 : 1;
Height of one satellite, h = 100 k.m

4R + 4h₂ = R + h₁ ⇒ h₁ = 3R + 4h₂. ;
Put h₂ = 100 km
∴ h₁ = 3 × 6400 + 400 = 19600 km.

Question 12.
A satellite is revolving round in a circular orbit with a speed of 8 km/ s^-1 at a height where the value of acceleration due to gravity is 8 m/s^-2. How high is the satellite from the Earth’s surface? (Radius of planet 6000 km.).
Solution:
Orbital velocity of satellite, V₀ = 8 km/s.
= 8 × 10³ m/s.
Acceleration due to gravity in the orbit = g
= 8 m/s²

Orbital velocity, V = \(\sqrt{gR}\)
where R is radius of the orbit and g is acceleration due to gravity in the orbit.

Height of satellite = 8000 – radius of earth ;
Radius of earth = 6000 km.
∴ Height above earth = 8000 – 6000
= 2000 km.

Question 13.
(a) Calculate the escape velocity of a body from the Earth’s surface, (b) If the Earth were made of wood, its mass would be 10% of its current mass. What would be the escape velocity, if the Earth were made of wood?
Solution:
Radius of earth, R = 6400 km = 6.4 × 10⁶ m.
Mass of earth, M = 6 × 10²⁴ kg; g = 9.8 ms^-2.
a) Escape velocity, V_e = \(\sqrt{2gR}\)
∴ V_e = \(\sqrt{2\times9.8\times6.4\times10^6}\) = 11.2 km/s

b) If earth is made of wood mass,
M₁ = 10% of M = 6 × 10²³

Additional Problems

Question 1.
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed (b) angular speed (c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
A comet while going on elliptical orbit around the Sun has constant angular momentum and total energy at all locations but other quantities vary with locations.

Question 2.
A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.5 × 10⁸ km away from the sun? Solution:
Here, T_s = 29.5 T_e; R_e = 1.5 × 10⁸ km; R_s = ?

Question 3.
A body weighs 63 N on the surface of Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Solution:
Weight of body = mg = 63 N
At height h, the value of g’ is given by,

Question 4.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of earth if it weighed 250 N on the surface?
Solution:
Weight of body at a depth, d = mg’