Gravitational Potential Energy

Practice Questions:

Question 1: easy

Gravitational potential difference between a point on surface of planet and another point \(10\text{ m}\) above is \(4\text{ J/kg}\). Considering gravitational field to be uniform, how much work is done in moving a mass of \(2.0\text{ kg}\) from the surface to a point \(5.0\text{ m}\) above the surface?

1. 0.40 J

2. 2.5 J

3. 4.0 J

4. 8.0 J

View Answer

For a uniform field, potential difference is proportional to distance. Thus, \(\Delta V' = \frac{5}{10} \times 4 = 2\text{ J/kg}\). The work done is \(W = m \Delta V' = 2.0 \times 2 = 4.0\text{ J}\).

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Question 2: easy

If escape velocity from earth is \(11.2\text{ km/s}\), Then escape velocity from a planet of mass as that of earth but of its one fourth radius

1. 11.2 km/s

2. 22.4 km/s

3. 5.6 km/s

4. 44.8 km/s

View Answer

Escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\). Since the mass of the planet is equal to that of Earth but the radius is \(R/4\), the escape velocity will be \(v'_e = \sqrt{\frac{2GM}{R/4}} = 2v_e = 2 \times 11.2 = 22.4\text{ km/s}\).

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Question 3: easy

A tunnel is dug along the diameter of the earth (radius \(R\) and mass \(M\)). There is a particle of mass \(‘m’\) at the centre of the tunnel. The minimum velocity given to the particle so that it just reaches to the surface of the earth is:

1. \(\sqrt{\frac{GM}{R}}\)

2. \(\sqrt{\frac{GM}{2R}}\)

3. \(\sqrt{\frac{2GM}{R}}\)

4. it will reach with the help of negligible velocity

View Answer

By conservation of mechanical energy, \(K_{\text{centre}} + U_{\text{centre}} = K_{\text{surface}} + U_{\text{surface}}\). With \(K_{\text{surface}} = 0\), we get \(\frac{1}{2}mv^2 - \frac{3GmM}{2R} = -\frac{GmM}{R}\), which gives \(v = \sqrt{\frac{GM}{2R}}\).

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Question 4: easy

What is the increase in gravitational potential energy of an object of mass \(m\) raised from the surface of earth to a height equal to \(n\) times of earth radius ?

1. \(\left(\frac{n+1}{n}\right) mgR\)

2. \(\left(\frac{n-1}{n}\right) mgR\)

3. \(\left(\frac{n}{n-1}\right) mgR\)

4. \(\left(\frac{n}{n+1}\right) mgR\)

View Answer

The increase in potential energy is \(\Delta U = U_f - U_i = -\frac{GMm}{R + nR} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} \left(1 - \frac{1}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR\).

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Question 5: easy

The gravitational potential energy of a body of mass \(m\) at the earth’s surface is \(-mgR_e\). Its gravitational potential energy at a height \(R_e\) from the earth’s surface will be (Here \(R_e\) is the radius of the earth)

1. \(-2mgR_e\)

2. \(2mgR_e\)

3. \(\frac{mgR_e}{2}\)

4. \(\frac{-mgR_e}{2}\)

View Answer

At the surface, \(U_s = -\frac{GMm}{R_e} = -mgR_e\). At a height \(h = R_e\), the distance from the center is \(r = 2R_e\). Thus, \(U_h = -\frac{GMm}{2R_e} = \frac{-mgR_e}{2}\).

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Question 6: easy

If mass of a planet is \(M\) and radius is \(x\), then the work to be done to slowly take a mass \(m\) from surface of planet to a height \(4x\) is :

1. \(\frac{4GMm}{5x}\)

2. \(\frac{2GMm}{5x}\)

3. \(\frac{5GMm}{x}\)

4. \(\frac{4GMm}{3x}\)

View Answer

Initial position \(r_i = x\), final position \(r_f = x + 4x = 5x\). Work done \(W = U_f - U_i = -\frac{GMm}{5x} - \left(-\frac{GMm}{x}\right) = \frac{4GMm}{5x}\).

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Question 7: easy

The escape velocity from the Earth’s surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is

1. \(4v\)

2. \(v\)

3. \(2v\)

4. \(3v\)

View Answer

Escape velocity is given by \(v_e = R\sqrt{\frac{8\pi G\rho}{3}}\). Since the mass density \(rho\) is the same, \(v_e\) is directly proportional to radius \(R\). Therefore, \(v_e' = 4v\).

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Question 8: easy

A particle of mass \(m\) is projected with a velocity \(v = k V_e\) (\(k < 1\)) from the surface of the earth. (\(V_e = \text{escape velocity}\)) The maximum height above the surface reached by the particle is

1. \(\frac{R k^2}{1-k^2}\)

2. \(R\left(\frac{k}{1-k}\right)^2\)

3. \(R\left(\frac{k}{1+k}\right)^2\)

4. \(\frac{R k^2}{1+k}\)

View Answer

By conservation of mechanical energy: \(-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h}\). Since \(v = k \sqrt{\frac{2GM}{R}}\), we substitute to get \(-\frac{1}{R}(1 - k^2) = -\frac{1}{R+h}\), which yields \(h = \frac{R k^2}{1-k^2}\).

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Question 9: easy

Two masses each equal to \(M\) are moving on a circular path of radius \(R\) about another fixed mass \(M\) (at the centre of the circular path). The gravitational potential energy of the system is:

1. \(-\frac{GM^2}{2R}\)

2. \(-\frac{GM^2}{R}\)

3. \(-\frac{2GM^2}{R}\)

4. \(-\frac{5GM^2}{2R}\)

View Answer

The total GPE of the three-mass system is \(U = -\frac{GMM}{R} - \frac{GMM}{R} - \frac{GMM}{2R} = -\frac{5GM^2}{2R}\) since the outer masses are at a distance of \(2R\) from each other and \(R\) from the center.

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Question 10: easy

If potential energy is assumed to be zero at infinity, then

1. The total energy of an orbiting satellite is negative of its potential energy.

2. The potential energy of an orbiting satellite is twice of its total energy

3. The potential energy of an orbiting satellite is negative of its kinetic energy

4. The total energy of an orbiting satellite is twice of its kinetic energy

View Answer

For an orbiting satellite, Potential Energy \(U = -\frac{GMm}{r}\), Kinetic Energy \(K = \frac{GMm}{2r}\), and Total Energy \(E = -\frac{GMm}{2r}\). This shows that \(U = 2E\).

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