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, \(U_h = -\frac{GMm}{2R_e} = \frac{-mgR_e}{2}\).
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 \(h = \frac{R k^2}{1-k^2}\).
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 \(U = 2E\).
Assertion (A): Gravitational potential energy of any mass particle may not be zero at earth centre.
Reason (R): Gravitational field intensity at earth centre is zero.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true. Gravitational potential at the center of a uniform solid sphere is finite (e.g., \( -\frac{3}{2} \frac{GM}{R} \)), hence potential energy is non-zero. Reason (R) is true. Due to symmetry, the net gravitational field at the Earth's center is zero. However, zero field does not directly explain non-zero potential energy. Thus, (R) is not the correct explanation of (A).
Assertion (A): If the product of surface area and density is same for two planets, escape velocities at surface will be same for both planets.
Reason (R): For given mass of a planet \( v_e \propto R^{-1/2} \)
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true.
Escape velocity \( v_e = \sqrt{\frac{2GM}{R}} \). Substitute \( M = \rho \frac{4}{3} \pi R^3 \) to get \( v_e = \sqrt{\frac{8G\pi R^2 \rho}{3}} \). If \( R^2 \rho \) is constant (derived from \( A\rho \) being constant), then \( v_e \) is constant. Reason (R) is true, as \( v_e = \sqrt{\frac{2GM}{R}} \) shows \( v_e \propto R^{-1/2} \) for constant \( M \). However, (R) does not explain (A), as the conditions are different.
Assertion (A): Escape velocity of a satellite is greater than its orbital velocity.
Reason (R): Orbit of a satellite is within the gravitational field of planet whereas escaping is beyond the gravitational field of planet.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true: escape velocity \(V_e = \sqrt{2GM/r}\) is \(\sqrt{2}\) times orbital velocity \(V_o = \sqrt{GM/r}\) for a circular orbit.
Reason (R) is false because the gravitational field extends infinitely. Escaping means overcoming the gravitational potential, not leaving the field.