Updated By SaraNextGen

On April 21, 2024, 11:35 AM

Question 8.11:

Choose the correct Answer from among the given ones:

For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Answer:

Answer: (ii)

Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient  is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.

If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

Question 8.12:

A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 ×10³⁰ kg, mass of the earth = 6 × 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 10¹¹ m).

Answer:

Mass of the Sun, M_s = 2 × 10³⁰ kg

Mass of the Earth, M_e = 6 × 10 ²⁴ kg

Orbital radius, r = 1.5 × 10¹¹ m

Mass of the rocket = m

Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.

From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:

Question 8.13:

How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 10⁸ km.

Answer:

Orbital radius of the Earth around the Sun, r = 1.5 × 10¹¹ m

Time taken by the Earth to complete one revolution around the Sun,

T = 1 year = 365.25 days

= 365.25 × 24 × 60 × 60 s

Universal gravitational constant, G = 6.67 × 10^–11 Nm² kg^–2

Thus, mass of the Sun can be calculated using the relation,

Hence, the mass of the Sun is 2 × 10³⁰ kg.

Question 8.14:

A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 ×10⁸ km away from the sun?

Answer:

Distance of the Earth from the Sun, r_e = 1.5 × 10⁸ km = 1.5 × 10¹¹ m

Time period of the Earth = T­e

Time period of Saturn, Ts = 29. 5 Te

Distance of Saturn from the Sun = r_s

From Kepler’s third law of planetary motion, we have

For Saturn and Sun, we can write

Hence, the distance between Saturn and the Sun is .

Question 8.15:

A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Answer:

Weight of the body, W = 63 N

Acceleration due to gravity at height h from the Earth’s surface is given by the relation:

Where,

g = Acceleration due to gravity on the Earth’s surface

Re = Radius of the Earth

Weight of a body of mass m at height h is given as:

Question 8.16:

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

Answer:

Weight of a body of mass m at the Earth’s surface, W = mg = 250 N

Body of mass m is located at depth, 

Where,

= Radius of the Earth

Acceleration due to gravity at depth g (d) is given by the relation:

Weight of the body at depth d,

Question 8.17:

A rocket is fired vertically with a speed of 5 km s^–1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G= 6.67 × 10^–11 N m² kg–^2.

Answer:

Answer: 8 × 10⁶ m from the centre of the Earth

Velocity of the rocket, v = 5 km/s = 5 × 10³ m/s

Mass of the Earth,

Radius of the Earth,

Height reached by rocket mass, m = h

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

At highest point h,

Total energy of the rocket

From the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface = Total energy at height h

Height achieved by the rocket with respect to the centre of the Earth

Question 8.18:

The escape speed of a projectile on the earth’s surface is 11.2 km s^–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Answer:

Escape velocity of a projectile from the Earth, v_esc = 11.2 km/s

Projection velocity of the projectile, v_p = 3v_esc

Mass of the projectile = m

Velocity of the projectile far away from the Earth = v_f

Total energy of the projectile on the Earth 

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth = 

From the law of conservation of energy, we have

Question 8.19:

A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 ×10²⁴ kg; radius of the earth = 6.4 ×10⁶ m; G = 6.67 × 10^–11 N m² kg^–2.

Answer:

Mass of the Earth, M = 6.0 × 10²⁴ kg

Mass of the satellite, m = 200 kg

Radius of the Earth, R_e = 6.4 × 10⁶ m

Universal gravitational constant, G = 6.67 × 10^–11 Nm²kg^–2

Height of the satellite, h = 400 km = 4 × 10⁵ m = 0.4 ×10⁶ m

Total energy of the satellite at height h

Orbital velocity of the satellite, v = 

Total energy of height, h

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.

Energy required to send the satellite out of its orbit = – (Bound energy)

Question 8.20:

Two stars each of one solar mass (= 2× 10³⁰ kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

Answer:

Mass of each star, M = 2 × 10³⁰ kg

Radius of each star, R = 10⁴ km = 10⁷ m

Distance between the stars, r = 10⁹ km = 10¹²m

For negligible speeds, v = 0 total energy of two stars separated at distance r

Now, consider the case when the stars are about to collide:

Velocity of the stars = v

Distance between the centers of the stars = 2R

Total kinetic energy of both stars 

Total potential energy of both stars 

Total energy of the two stars = 

Using the law of conservation of energy, we can write:

Question 8.21:

Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Answer:

Answer:

0;

–2.7 × 10^–8 J /kg;

Yes;

Unstable

Explanation:

The situation is represented in the given figure:

Mass of each sphere, M = 100 kg

Separation between the spheres, r = 1m

X is the mid point between the spheres. Gravitational force at point X will be zero. This is because gravitational force exerted by each sphere will act in opposite directions.

Gravitational potential at point X:

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

Question 8.22:

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 10²⁴ kg, radius = 6400 km.

Answer:

Mass of the Earth, M = 6.0 × 10²⁴ kg

Radius of the Earth, R = 6400 km = 6.4 × 10⁶ m

Height of a geostationary satellite from the surface of the Earth,

h = 36000 km = 3.6 × 10⁷ m

Gravitational potential energy due to Earth’s gravity at height h,

Question 8.23:

A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun = 2 × 10³⁰ kg).

Answer:

Answer: Yes

A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, f_g

Where,

M = Mass of the star = 2.5 × 2 × 10³⁰ = 5 × 10³⁰ kg

m = Mass of the body

R = Radius of the star = 12 km = 1.2 ×10⁴ m

Centrifugal force, f_c = mrω² ω = Angular speed = 2πν

ν = Angular frequency = 1.2 rev s^–1

f_c = mR (2πν)²

= m × (1.2 ×10⁴) × 4 × (3.14)² × (1.2)² = 1.7 ×10⁵m N

Since f_g > f_c, the body will remain stuck to the surface of the star.

Question 8.24:

A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 10³⁰ kg; mass of mars = 6.4 × 10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10⁸kg; G= 6.67 × 10^–11 m²kg^–2.

Answer:

Mass of the spaceship, m_s = 1000 kg

Mass of the Sun, M = 2 × 10³⁰ kg

Mass of Mars, mm = 6.4 × 10 ²³ kg

Orbital radius of Mars, R = 2.28 × 10⁸ kg =2.28 × 10¹¹m

Radius of Mars, r = 3395 km = 3.395 × 10⁶ m

Universal gravitational constant, G = 6.67 × 10^–11 m²kg^–2

Potential energy of the spaceship due to the gravitational attraction of the Sun 

Potential energy of the spaceship due to the gravitational attraction of Mars

Since the spaceship is stationed on Mars, its velocity and hence, its kinetic energy will be zero.

Total energy of the spaceship 

The negative sign indicates that the system is in bound state.

Energy required for launching the spaceship out of the solar system

= – (Total energy of the spaceship)

Question 8.25:

A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10^-11 N m² kg^–2.

Answer:

Initial velocity of the rocket, v = 2 km/s = 2 × 10³ m/s

Mass of Mars, M = 6.4 × 10²³ kg

Radius of Mars, R = 3395 km = 3.395 × 10⁶ m

Universal gravitational constant, G = 6.67× 10^–11 N m² kg^–2

Mass of the rocket = m

Initial kinetic energy of the rocket = 

Initial potential energy of the rocket

Total initial energy

If 20 % of initial kinetic energy is lost due to Martian atmospheric resistance, then only 80 % of its kinetic energy helps in reaching a height.

Total initial energy available 

Maximum height reached by the rocket = h

At this height, the velocity and hence, the kinetic energy of the rocket will become zero.

Total energy of the rocket at height h 

Applying the law of conservation of energy for the rocket, we can write: