This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Gravitation – Escape Speed”.

1. The ratio of the radius of a planet A to that of a planet B is “r”. The ratio of acceleration due to gravity on the planets is “p”. The ratio of escape velocities of the two planets is _____

a) (pr)^{1/2}

b) (p/r)^{1/2}

c) (pr)

d) (p/r)

View Answer

Explanation: Escape velocity (v) = (2*g*R)

^{1/2}

g = Acceleration due to gravity

R = Radius of planet

Given;

R

_{a}/R

_{b}= r

g

_{a}/g

_{b}= p

v

_{a}/v

_{b}= (2*g

_{a}*R

_{a}])

^{1/2}/(2*g

_{b}*R

_{b})

^{1/2}

= (p x r)

^{1/2}.

2. What is the minimum velocity required for an object of mass “m” to escape the gravitational pull of a planet of mass “M” and radius “R” from its surface?

a) [(G*M)/R]^{1/2}

b) [(2*G*M)/R] ^{2}

c) [(G*M*m)/R]^{1/2}

d) [(2*G*M)/R]^{1/2}

View Answer

Explanation: If the object’s initial velocity is “v”, from the law of conservation of energy, we have;

(m x v

^{2})/2 = (G x M x m)/R

v

^{2}= (2 x G x M)/R

Therefore; v = [(2 x G x M)/R]

^{1/2}.

3. The escape velocity of an object from the surface of the planet does not depend on the mass of the object.

a) True

b) False

View Answer

Explanation: The escape velocity of an object depends only on the mass of the planet from which it is escaping and not the mass of the object. The relation is as given below;

Escape velocity (v) = [(2*G*M)/R]

^{1/2}

R = Radius of planet

M = Mass of planet.

4. An object is projected vertically upwards with a velocity (gr) ^{1/2}. What is the maximum height reached by the object if “R” is the radius of the earth and “g” is the acceleration due to gravity?

a) R/2

b) R

c) 2R

d) 4R

View Answer

Explanation: From the law of conservation of energy, we have;

(m x v

^{2})/2 – (G x M x m)/R = -(G x M x m)/(R+h)

m = Mass of object

M = Mass of earth

h = Maximum height of the object

v = Velocity of projection

(m x v

^{2})/2 – (g x m x R) = -(g x m x R

^{2})/(R+h)

(g x R)/2 = (g x R x h)/(R+h)

Therefore; h = R.

5. The escape velocity of the earth is “v”. If an object is thrown vertically upwards with a velocity “kv”, what is the speed of the object at infinity?

a) v/(k^{2}-1)^{1/2}

b) v(k^{2}-1)^{1/2}

c) v^{2}/(k^{2}-1)

d) v^{2} (k^{2}-1)

View Answer

Explanation: From the law of conservation of energy, we have;

(m x (kv)

^{2})/2 – (G x M x m)/R = (m x p

^{2})/2

m = Mass of object

M = Mass of earth

R = Radius of the earth

p = Velocity of the object at infinity

(kv)

^{2}/2 –(g x R) = p

^{2}/2

(kv)

^{2}/2 – v

^{2}/2 = p

^{2}/2; v = (2 x g x R)

^{½}

(k

^{2}-1)v

^{2}= p

^{2}

Therefore; p = v(k

^{2}-1)

^{1/2}.

6. The earth is able to retain its atmosphere because of _____

a) the mean velocities of atmospheric gas molecules which is less than that of the earth’s escape velocity

b) the mean velocities of atmospheric gas molecules which is greater than that of the earth’s escape velocity

c) the gravitational effect of the moon

d) the earth’s magnetic field

View Answer

Explanation:The escape velocity of earth is approximately 11.2km/s. This is greater than the root-mean-square velocities of the atmospheric gases. Hence, the gases are unable to escape into the vacuum of space.

7. The escape velocity at the event horizon of a black hole is 3×10^{8} m/s, i.e., the velocity of light. What is the mass of the black hole if the distance from its centre to the event horizon is 18km?

a) 12.15 x 10^{-30} kg

b) 12.15 x 10^{30} kg

c) Infinity

d) Cannot be determined

View Answer

Explanation: Escape velocity (v) = [(2 x G x M)/R]

^{1/2}

M = Mass of the black hole

R = Radius of the black hole

3 x 10

^{8}= [(2 x 6.67 x 10

^{-11}x M)/18000]

^{1/2}

9 x 10

^{16}= (2 x 6.67 x 10

^{-11}x M)/18000

M = 12.15 x 10

^{30}kg.

8. A black hole is called so because it is a hypothetical object.

a) True

b) False

View Answer

Explanation:A black hole is called so because of its high escape velocity. A black holes escape velocity is equal to the velocity of light at the event horizon, and hence, it absorbs any light hitting it without any reflection which essentially renders the object “black”.

9. The radius of the moon is approximately 3.7 times smaller than the radius of the earth. Assuming that their densities are same, what is the escape velocity of the moon compared to earth?

a) 0.07 times

b) 0.7 times

c) 7 times

d) 3.7 times

View Answer

Explanation: Escape velocity (v) = [(2*G*M)/R]

^{1/2}

Mass of moon = (Density) x (Volume)

= (Density) x (Volume of the earth)/50.65;

[Volume is directly proportional to the cube of the radius]

V

_{moon}/v

_{earth}= [(2*G*M

_{moon})/R

_{moon}

^{1/2}/[(2*G*M

_{earth})/R

_{earth}]

^{1/2}

= [1/(50.65 x 3.7)]

^{½}

= 0.07

Therefore; V

_{moon}= 0.07 (v

_{earth}).

10. The escape velocity of a planet depends on the radius of its orbit around the parent star.

a)True

b)False

View Answer

Explanation:The escape velocity of a planet is independent of the radius of orbit of its parent star. It only depends on the radius of the planet and the mass of the planet.

11. The escape velocity for an object to leave the surface of the earth does not depend on its size, shape, mass or direction of projection of the body.

a) True

b) False

View Answer

Explanation: The escape velocity for any object on earth is approximately constant at all points on the earth’s surface, i.e., the escape velocity of earth is approximately equal to 11.2 km/s and is not determined by any parameter of the projectile.

12. A planet as a radius “R” and density “P”. The escape velocity of this planet is _____

a) directly proportional to P

b) inversely proportional to P

c) directly proportional to P^{1/2}

d) inversely proportional to P^{1/2}

View Answer

Explanation: Escape velocity (v) = [(2*G*M)/R]

^{1/2}

Density (P) = Mass (M) x Volume

M = P / [(4/3) x pi x R

^{3}] Therefore;

v = [(8 x pi x P x G x R

^{2}) / 3]

^{1/2}.

13. A particle is kept at a distance R above the earth’s surface. What is the minimum speed with which it should be projected so that it does not return?

a) [(G*M)/(4*R)]^{1/2}

b) [(G*M)/(2*R)]^{1/2}

c) [(G*M)/R] ^{1/2}

d) [(2*G*M)/R]^{1/2}

View Answer

Explanation:If the object’s initial velocity is “v”, from the law of conservation of energy, we have;

(m x v

^{2})/2 = (G x M x m)/(R+R)

v

^{2}= (2 x G x M)/(2x R)

Therefore; v = [(G x M)/R]

^{1/2}.

14. The radius of Jupiter is approximately 11 times larger than the earth. Jupiter has a mass 316 times that of earth. What is the approximate escape velocity of Jupiter compared to earth?

a) 100 times greater

b) 12 times greater

c) 5 times greater

d) Equal to that of the earth

View Answer

Explanation:M = Mass of the earth

R = Radius of the earth

v = Escape velocity of the earth

We know;

v = [(2 x G x M)/R]

^{1/2}

M’ = Mass of the Jupiter

R’ = Radius of the Jupiter

v’ = Escape velocity of the Jupiter

M’ = 316M

R’ = 11R

v’ = [(2 x G x M’)/R’]

^{½}

= [(2 x G x 316M)/11R]

^{½}

= 5.6 x [(2 x G x M)/R]

^{1/2}

= 5.6 x v.

15. If the earth lost 99% of its mass, the escape velocity would _____

a) increase 10 times

b) decrease 10 times

c) decrease 90 times

d) increase 90 times

View Answer

Explanation: New mass of the earth (M’) = 0.01 x Original mass of the earth (M)

Therefore;

M’ = M/100

Escape velocity (v) is directly proportional to the square root of the mass

“v” is directly proportional to the square root of “M’”.

Therefore;

New escape velocity (v’) = v/10.

**Sanfoundry Global Education & Learning Series – Physics – Class 11**.

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