A $10 k g$ satellite completes one revolution around the earth at a height of $100 k m$ in $108 m i n u t e s$ .The work done by the gravitational force of earth will be:

108 \times 100 \times 10 J

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\frac{108 \times 10}{100} J

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0 J

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\frac{100 \times 10}{108} J

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Solution

The correct option is C $0 J$
$g r a v i t a t i o n a l f o r c e a c t s t o w a r d s c e n t r e o f e a r t h . d i s p l a c e m e n t t o w a r d s c e n t r e o f e a r t h = 0; a s d i s p l a c e m e n t i n t h e d i r e c t i o n o f f o r c e = 0. w o r k d o n e b y g r v i t a t i o n a l f o r c e = 0.$

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A 10 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes.The work done by the gravitational force of earth will be:

The correct option is C 0 Jgravitationalforceactstowardscentreofearth.displacementtowardscentreofearth=0;asdisplacementinthedirectionofforce=0.workdonebygrvitationalforce=0.

A $10 k g$ satellite completes one revolution around the earth at a height of $100 k m$ in $108 m i n u t e s$ .The work done by the gravitational force of earth will be: