What is the ratio of escape velocity at Earth to the escape velocity at a planet whose radius and mean density are twice that of Earth?

The correct ratio is 1 : 2√2. Escape velocity in terms of density and radius is directly proportional to the radius multiplied by the square root of the density. If both the radius and density are doubled for the planet, its escape velocity becomes 2√2 times that of Earth.

If the radius of Earth contracts to 1/4th of its value while keeping the mass constant, what will happen to the escape velocity?

The escape velocity will be doubled. Escape velocity is inversely proportional to the square root of the planet's radius. If the radius is reduced to 1/4, the square root of 1/4 is 1/2. Dividing the original equation by 1/2 results in the velocity doubling.

Case Study Questions for Class 11 Physics Chapter 8 Gravitation

Q: Gas escapes from the surface of a planet because it acquires an escape velocity. The escape velocity depends on which factors?

Escape velocity depends on the mass of the planet. The escape velocity formula relies entirely on the mass of the planet and its radius. It is completely independent of the mass of the escaping particle or gas.

Q: If a satellite with a very small mass compared to Earth is moving in an elliptical orbit around the Earth, what happens to its acceleration?

The acceleration of the satellite is always directed towards the centre of the Earth. The gravitational force between the Earth and the satellite provides the necessary centripetal acceleration, which always points toward the center of the Earth along the radius vector.

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Home CBSE Class 11 Physics Case Study Qs Ch 8 Gravitation

Here we are providing case study questions for class 11 physics chapter 8 Gravitation. Students can practice these questions for better understanding of case study type questions.

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Key Concepts at a Glance

Read before attempting the questions below

Ch 8 · Gravitation · Class 11

Universal Law of Gravitation Formula

Every particle attracts every other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

\( F = \dfrac{G \cdot m_1 \cdot m_2}{r^2} \)

Gravitational Constant G

G is a universal constant — it does not depend on the nature, size, or mass of the objects, nor on the medium between them.

G = 6.674 × 10⁻¹¹ N·m²·kg⁻²

Escape Velocity Formula

Minimum velocity needed to escape Earth’s gravity. Does not depend on the mass of the projected body.

vₑ = √(2gR) ≈ 11.2 km·s⁻¹

Case Study Questions for Class 11 Physics Chapter 8 Gravitation

3 Case Studies · 12 Questions · With Answer & Explanation

Case Study: Earth’s Satellite

Read the passage carefully, then answer all four questions

Ch 8 · Gravitation

Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately 27.3 days which is also roughly equal to the rotational period of the moon about its own axis.

Also, the speed that a satellite needs to be travelling to break free of a planet or moon’s gravity well and leave it without further propulsion is known as escape velocity. For example, a spacecraft leaving the surface of earth needs to be going 7 miles per second or nearly 25000 miles per hour to leave without falling back to the surface or falling into orbit.

The ratio of escape velocity at earth (v_e) to the escape velocity at a planet (v_p), whose radius and mean density are twice as that of earth is

a 1 : 2√2

b 1 : 4

c 1 : √2

d 1 : 2

Correct Answer (a) 1 : 2√2 ✅

Explanation

Escape velocity in terms of density (ρ) and radius (R) is directly proportional to R√(ρ). If both radius and density are doubled for the planet, v_p becomes 2√(2) times v_e.

v_e = R√(8πρG/3)

Gas escapes from the surface of a planet because it acquires an escape velocity. The escape velocity will depend on which of the following factors?

a Mass of the planet

b Mass of the particle escaping

c Temperature of the planet

d None of the above

Correct Answer (a) Mass of the planet ✅

Explanation

The escape velocity formula relies entirely on the mass of the planet (M) and its radius (R). It is completely independent of the mass of the escaping particle.

v_e = √(2GM/R)

iii

The escape velocity of a satellite from the earth is v_e. If the radius of earth contracts to (1/4)^th of its value, keeping the mass of the earth constant, escape velocity will be

a doubled

b halved

c tripled

d unaltered

Correct Answer (a) doubled ✅

Explanation

Escape velocity is inversely proportional to the square root of the radius. If R is reduced to R/4, the square root of 1/4 is 1/2. Dividing by 1/2 means the velocity is doubled.

v ∝ 1/√(R)

A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth, then

a the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant

b the total mechanical energy of S varies periodically with time

c the linear momentum of S remains constant in magnitude

d the acceleration of S is always directed towards the centre of the earth

Correct Answer (d) the acceleration of S is always directed towards the centre of the earth ✅

Explanation

The gravitational force between the Earth and the satellite provides the necessary centripetal acceleration, which is always directed toward the center of the Earth along the radius vector.

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Common Mistake

Students often confuse orbital velocity (8 km/s) with escape velocity (11.2 km/s). Remember — orbital velocity keeps a satellite in orbit; escape velocity allows it to leave Earth’s gravitational field entirely.

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