Ncert Exercises & Solutions

8.22 As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to the earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0×1024 kg, radius = 6400 km.

G i v e n,

M = 100 k g

r = 1 m

I n f i g u r e X i s t h e m i d p o i n t b e t w e e n t h e s p h e r e s . G r a v i t a t i o n a l f o r c e a t p o n t X w i l l b e z e r o .

(G r a v i t a t i o n a l f o r c e e x e r t e d b y e a c h s p h e r e a r e e q u a l a n d i n o p p o s i t e i r e c t i o n)

G r a v i t a t i o n a l p o t e n t i a l a t p o i n t X :

= - \frac{G M}{(r / 2)} - \frac{G M}{(r / 2)} = - 4 \frac{G M}{r}

= - \frac{4 \times 6.67 \times 10^{- 11} \times 100}{1}

= - 2.67 \times 10^{- 8} J k g^{- 1}

Any object placed at point X will be in an equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

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