Two bodies of masses \(m\) and \(M\) are placed at distance \(d\) apart. What is the gravitational potential \(V\) at the position where the gravitational field due to them is zero?
1. \(V = -\frac{G}{d}(m + M)\)
2. \(V = -\frac{G}{d} m\)
3. \(V = -\frac{GM}{d}\)
4. \(V = -\frac{G}{d}(\sqrt{m} + \sqrt{M})^2\)
View Answer
At the point where the field is zero, \(\frac{Gm}{r_1^2} = \frac{GM}{r_2^2}\), which gives \(r_1 = \frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d\) and \(r_2 = \frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d\). The potential is \(V = -\frac{Gm}{r_1} - \frac{GM}{r_2} = -\frac{G}{d}(\sqrt{m} + \sqrt{M})^2\).
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 \(W = m \Delta V' = 2.0 \times 2 = 4.0\text{ J}\).
Two concentric shells have mass \(M\) and \(m\) and their radii are \(R\) and \(r\) respectively, where \(R > r\). What is the gravitational potential at their common centre ?
1. \(-\frac{GM}{R}\)
2. \(-\frac{GM}{r}\)
3. \(-G \left[ \frac{M}{R} - \frac{m}{r} \right]\)
4. \(-G \left[ \frac{M}{R} + \frac{m}{r} \right]\)
View Answer
The gravitational potential inside any spherical shell is constant and equals the potential at its surface. Therefore, the total potential at the common centre is the sum of the potentials: \(V = -\frac{GM}{R} - \frac{Gm}{r} = -G\left[\frac{M}{R} + \frac{m}{r}\right]\).
The gravitational force between two particles with masses \(m\) and \(M\), initially at rest at great separation, pulls them together. When their separation becomes \(d\), then speed of either particle relative to the other will be :
1. \(\sqrt{G(M+m)/2d}\)
2. \(\sqrt{G(M+m)/d}\)
3. \(\sqrt{4G(M+m)/d}\)
4. \(\sqrt{2G(M+m)/d}\)
View Answer
By conservation of mechanical energy, the relative speed is found using the reduced mass \(\mu = \frac{mM}{m+M}\). Thus, \(\frac{1}{2} mu v_{\text{rel}}^2 = \frac{GMm}{d}\), which simplifies to \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).
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 \(v = \sqrt{\frac{GM}{2R}}\).
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 \(\Delta U = U_f - U_i = -\frac{GMm}{R + nR} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} \left(1 - \frac{1}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR\).
A satellite is launched in the equatorial plane in such a way that it can transmit signals upto \(60^\circ\) latitude on the earth. Then the angular velocity of the satellite is :
1. \(\sqrt{\frac{GM}{8R^3}}\)
2. \(\sqrt{\frac{GM}{2R^3}}\)
3. \(\sqrt{\frac{GM}{4R^3}}\)
4. \(\sqrt{\frac{3\sqrt{3}GM}{8R^3}}\)
View Answer
For a transmission coverage up to latitude \(\theta = 60^\circ\), the orbital radius \(r\) satisfies \(\cos\theta = \frac{R}{r}\). This gives \(r = 2R\). The angular velocity is \(\omega = \sqrt{\frac{GM}{r^3}} = \sqrt{\frac{GM}{8R^3}}\).
A satellite is seen after each 8 hours over equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth & satellite is same will be :
1. 8 hours
2. 12 hours
3. 24 hours
4. 6 hours
View Answer
When rotating oppositely, \(\frac{1}{T_{\text{rel}}} = \frac{1}{T_s} + \frac{1}{T_e} \Rightarrow \frac{1}{8} = \frac{1}{T_s} + \frac{1}{24}\), which gives \(T_s = 12\text{ hours}\). When rotating in the same direction, \(\frac{1}{T_{\text{rel}}'} = \frac{1}{T_s} - \frac{1}{T_e} = \frac{1}{12} - \frac{1}{24} = \frac{1}{24}\), so \(T_{\text{rel}}' = 24\text{ hours}\).