A body of mass m is situated at a distance eq | vedclass

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

Question

(B) The distance of the body from the center of the Earth is $r = R + 2R = 3R$.The gravitational potential energy $(PE)$ of the body at this distance

Vedclass · Accepted Answer

$\frac{mgR}{3}$

Vedclass · Answer

(B) The distance of the body from the center of the Earth is $r = R + 2R = 3R$.The gravitational potential energy $(PE)$ of the body at this distance is given by $PE = -\frac{GMm}{r} = -\frac{GMm}{3R}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the potential energy expression: $PE = -\frac{(gR^2)m}{3R} = -\frac{mgR}{3}$.The energy required to escape the gravitational field is equal to the magnitude of the potential energy,which is the binding energy: $E_{escape} = -PE = \frac{mgR}{3}$.

$A$ body of mass $m$ is situated at a distance equal to $2R$ ($R$ is the radius of the Earth) from the Earth's surface. The minimum energy required to be given to the body so that it may escape out of the Earth's gravitational field is:

Similar Questions

The escape velocity of a planet having mass $6$ times and radius $2$ times as that of Earth is

The kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$) to infinity is:

$A$ body is projected up with a velocity equal to $\frac{3}{4}$ of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth $= R$)

$A$ body of mass $m$ is situated at a distance $4R_e$ above the Earth's surface,where $R_e$ is the radius of Earth. How much minimum energy must be given to the body so that it may escape?

There is no atmosphere on the moon because

Vedclass Test Series

Exam Paper Generator

Online Exam Module