NEET Gravitational Potential Energy MCQs & PYQs

Practice NEET Gravitational Potential Energy Questions

Question 1:

moderate

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 $.

💬 Discuss this Question

Question 2:

moderate

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 $v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}$.

💬 Discuss this Question

Question 3:

moderate

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 $\Delta U = \frac{mg(nR)}{1 + n} = \left(\frac{n}{n+1}\right) mgR$.

💬 Discuss this Question

Question 4:

moderate

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 $v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}$.

💬 Discuss this Question

Question 5:

moderate

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 $v^2 = \frac{3GM}{R} = 1.5 v_e^2⇒ v = \sqrt{1.5} v_e$.

💬 Discuss this Question

Question 6:

moderate

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 $E = 6.4 \times 10^{10}\text{ Joules}$.

💬 Discuss this Question

Question 7:

moderate

An artificial satellite is moving in a circular orbit of radius $r$ around a planet. Total energy of satellite is $E$. Energy required to move this satellite into a new orbit of radius $2r$, is

The total energy in orbit is $E = -\frac{GMm}{2r}$. In the new orbit of radius $2r$, total energy is $E' = -\frac{GMm}{4r} = \frac{E}{2}$. The required energy is $\Delta E = E' - E = \frac{E}{2} - E = -\frac{E}{2}$.

💬 Discuss this Question

Question 8:

moderate

A rocket is fired vertically with a speed half of the escape speed from the earth’s surface. How far from the earth does the rocket go before returning to the earth? [Given mean radius of earth is $R$]

Using conservation of energy: $-\frac{GMm}{R} + \frac{1}{2}m v^2 = -\frac{GMm}{r}$. Putting $v = \frac{v_e}{2} = \sqrt{\frac{GM}{2R}}$ yields $r = \frac{4R}{3}$, which is the distance from the earth's centre.

💬 Discuss this Question

Question 9:

moderate

The work done to raise a mass $m$ from the surface of the earth to a height $h$, which is equal to the radius of the earth, is:

(2019)

Work done $W = \Delta U = \frac{mgh}{1+h/R}$. Substituting $h = R$, we get $W = \frac{mgR}{1+1} = \frac{1}{2}mgR$.

💬 Discuss this Question

Question 10:

moderate

A body of mass ‘$m$’ taken from the earth’s surface to the height equal to twice the radius ($R$) of the earth. The change in potential energy of body will be:

(2013)

Change in potential energy $\Delta U = \frac{mgh}{1+h/R}$. For $h = 2R$, $\Delta U = \frac{mg(2R)}{1+2} = \frac{2}{3}mgR$.

💬 Discuss this Question