If the acceleration due to gravity at the surface of the earth is g ,the work done in slowly lifting - Physics - Gravitation

Question

If the acceleration due to gravity at the surface of the earth
is g ,the work done in slowly lifting a body of mass m from the
earths surface to a height R equal to the radius of the earth
is
a)1/2mgr
b)2mgr
c)mgr
d)none of these

Vijay Yadav · Accepted Answer

Hello, Here force applied is variable because acceleration due to gravity change with height $So g = \frac{GM}{(R+ h)^{2}}$ {h = height from earth surface which is variable} So force applied $F = mg = \frac{GM m}{(R+ h)^{2}}$

Small work done by applied force to lift height dh $d W= \vec{F} \cdot d \vec{h} S o t o t a l w o r k d o n e i s g i v e n b y \Rightarrow W= \int_{0}^{R} d W = \int_{0}^{R} \vec{F} \cdot d \vec{h} = \int_{0}^{R} F d h \cos 0 ° = \int_{0}^{R} F d h = \int_{0}^{R} \frac{GM m}{(R+ h)^{2}} d h \Rightarrow W= (GM m) \int_{0}^{R} \frac{d h}{(R+ h)^{2}} = (GM m) {[- \frac{1}{(R+ h)}]_{0}^{R}}^{} = GM m [- \frac{1}{R+R} + \frac{1}{R}] = GM m [- \frac{1}{2 R} + \frac{1}{R}] = \frac{GM m}{2 R} = \frac{1}{2} (\frac{GM}{R^{2}}) \times m \times R$ $W= \frac{1}{2} gmR = \frac{1}{2} mgR$ So option (a) is correct

If the acceleration due to gravity at the surface of the earth is g ,the work done in slowly lifting a body of mass m from the earths surface to a height R equal to the radius of the earth is. a)1/2mgr b)2mgr c)mgr d)none of these

If the acceleration due to gravity at the surface of the earth is g ,the work done in slowly lifting a body of mass m from the earths surface to a height R equal to the radius of the earth is.

a)1/2mgr

b)2mgr

c)mgr

d)none of these