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
Using energy conservation in the center-of-mass frame: \(\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{GMm}{d}\), where \(\mu = \frac{Mm}{M+m}\) is the reduced mass. Substituting \(\mu\) yields \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).
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 \(\Delta U = \frac{mgh}{1 + h/R}\). Since \(h = nR\), we get \(\Delta U = \frac{mg(nR)}{1 + n} = \left(\frac{n}{n+1}\right) mgR\).
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 \(W = m \Delta V' = 2.0 \times 2 = 4.0\text{ J}\).
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 \(\Delta U = U_f - U_i = -\frac{GMm}{R + nR} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R} \left(1 - \frac{1}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR\).
The escape velocity for a planet is \(v_e\). A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be
1. \(\sqrt{1.5} v_e\)
2. \(\frac{v_e}{\sqrt{2}}\)
3. \(v_e\)
4. zero
View Answer
Using conservation of mechanical energy from infinity to the centre: \(0 = \frac{1}{2}mv^2 - \frac{3GmM}{2R}\). Since escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\), we get \(v^2 = \frac{3GM}{R} = 1.5 v_e^2⇒ v = \sqrt{1.5} v_e\).
The mass of a spaceship is \(1000\text{ kg}\). It is to be launched from the earth’s surface out into free space. The value of \(‘g’\) and \(‘R’\) (radius of earth) are \(10\text{ m/s}^2\) and \(6400\text{ km}\) respectively. The required energy for this work will be :
1. \(6.4 \times 10^{10}\text{ Joules}\)
2. \(6.4 \times 10^{11}\text{ Joules}\)
3. \(6.4 \times 10^8\text{ Joules}\)
4. \(6.4 \times 10^9\text{ Joules}\)
View Answer
The minimum energy required to escape the earth's gravitational pull from the surface is \(E = \frac{GMm}{R} = mgR\). Substituting \(m = 1000\text{ kg}\), \(g = 10\text{ m/s}^2\), and \(R = 6.4 \times 10^6\text{ m}\), we find \(E = 6.4 \times 10^{10}\text{ Joules}\).