The change in the gravitational potential ene | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.At the surface of the Earth

Vedclass · Accepted Answer

$\left(\frac{n}{n+1}\right) mgR$

Vedclass · Answer

(A) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.At the surface of the Earth, $r = R$, so $U_i = -\frac{GMm}{R}$.At a height $h = nR$ above the surface, the distance from the center is $r = R + nR = R(1+n)$.So, $U_f = -\frac{GMm}{R(1+n)}$.The change in gravitational potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{R(1+n)} - (-\frac{GMm}{R})$.$\Delta U = \frac{GMm}{R} \left(1 - \frac{1}{1+n}\right) = \frac{GMm}{R} \left(\frac{1+n-1}{1+n}\right) = \frac{GMm}{R} \left(\frac{n}{1+n}\right)$.Since $g = \frac{GM}{R^2}$, we have $GM = gR^2$.Substituting this, $\Delta U = \frac{gR^2 m}{R} \left(\frac{n}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR$.

The change in the gravitational potential energy when a body of mass $m$ is raised to a height $h = nR$ above the surface of the Earth is (where $R$ is the radius of the Earth).

Similar Questions

$A$ body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h = 3R$ ($R$ is the radius of the earth). The change in gravitational potential energy of the body is

The energy required to take a body from the surface of the earth to a height equal to the radius of the earth is $W$. The energy required to take this body from the surface of the earth to a height equal to twice the radius of the earth is:

If the gravitational acceleration at the surface of the Earth is $g$,then the increase in potential energy in lifting an object of mass $m$ to a height equal to half of the radius of the Earth from the surface will be:

$A$ body starts from rest from a distance $R_0$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ($R=$ radius of earth,$M=$ mass of earth).

$A$ body which is initially at rest at a height $R$ above the surface of the earth of radius $R$,falls freely towards the earth. Its velocity on reaching the surface of the earth is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module