A body is to be projected vertically upwards from Earth rsquo-s surface to reach a height of 9R where R is the radius of the Earth- What is the velocity required to do so-

Let #160;the #160;velocity #160;of #160;the #160;body #160;be #160;v #160;and #160;height #160;reached #160;is #160;9R.So #160; #160 ...

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Standard IX

Physics

Acceleration Due to Gravity

A body is to ...

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A body is to be projected vertically upwards from Earth’s surface to reach a height of 9R where R is the radius of the Earth. What is the velocity required to do so?

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$L e t t h e v e l o c i t y o f t h e b o d y b e v a n d h e i g h t r e a c h e d i s 9 R . S o k i n e t i c e n e r g y o f p r o j e c t i o n = P . E . a t h e i g h t 9 R \frac{1}{2} m v^{2} = m g \times 9 R o r v^{2} = 18 g R o r v = 3 \sqrt{2 g R} = 3 \sqrt{\frac{2 G M}{R}}$

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A body is to be projected vertically upwards from Earth’s surface to reach a height of 9R where R is the radius of the Earth. What is the velocity required to do so?

Let the velocity of the body be v and height reached is 9R.So kinetic energy of projection=P. E. at height 9R 12mv2=mg×9R or v2=18gRor v=32gR=32GMR

$L e t t h e v e l o c i t y o f t h e b o d y b e v a n d h e i g h t r e a c h e d i s 9 R . S o k i n e t i c e n e r g y o f p r o j e c t i o n = P . E . a t h e i g h t 9 R \frac{1}{2} m v^{2} = m g \times 9 R o r v^{2} = 18 g R o r v = 3 \sqrt{2 g R} = 3 \sqrt{\frac{2 G M}{R}}$