Gravitational Potential Energy - NEET Physics Chapterwise MCQs & PYQs
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NEET Gravitational Potential Energy MCQs & PYQs
Practice NEET Gravitational Potential Energy Questions
Question 1:
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The ratio of the radius of the earth to that of the moon is 10. The ratio of acceleration due to gravity on the earth and on the moon is 6. The ratio of the escape velocity from the earth’s surface to that from the moon is:
Escape velocity is given by the formula \( v_e = \sqrt{2gR} \). The ratio of escape velocity of earth to moon is \( \frac{v_{earth}}{v_{moon}} = \sqrt{\frac{g_{earth}}{g_{moon}} \times \frac{R_{earth}}{R_{moon}}} = \sqrt{6 \times 10} = \sqrt{60} \approx 7.75 \approx 8 \).
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 :
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 ?
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?
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 :
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 \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).
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
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 \(v'_e = \sqrt{\frac{2GM}{R/4}} = 2v_e = 2 \times 11.2 = 22.4\text{ km/s}\).
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:
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 ?
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
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 :
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 \(E = 6.4 \times 10^{10}\text{ Joules}\).