Q. An object weighs 10N at the north-pole of the earth. In a geostationary satellite distance $7R$ from the centre of earth (of radius $R$) what will be its true weight?
A) 3 N
B) 5 N
C) 2 N
D) 0.2 N
Explanation
(d) The true weight of a body is given by $mg$ and with height $g$ decrease
So, $\frac{W_{s}}{W_{E}} = \frac{mg'}{mg} = \frac{1}{\lbrack 1 + {(h/R)\rbrack}^{2}}\left\lbrack as\ g^{'} = \frac{g}{\left\lbrack 1 + (h/R) \right\rbrack^{2}} \right\rbrack$
But here, $h = 7R - R = 6R,\ \ ie,\ h/R = 6$
So, $W_{S} = \frac{W_{E}}{(1 + 6)^{2}} = \frac{10}{49} = 0.2N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.