Question

the escape velocity of the a body on the surface is 11.2km/sec if the earth's mass increase to twice it's present value and redius of the earth becomes half the escape velocity become

Answer 1

☆☆AS-SALAMU-ALYKUM ☆☆

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ESCAPE VELOCITY IS GIVEN BY :-
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Vesc= SQUARE ROOT 2 GM÷R

ESCAPE VELOCITY OF EARTH IS 11.2km/sec

GIVEN DATA :-
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MASS IS INCREAED TO TWICE =M' = 2M

RADIUS IS HALVED = R' =R÷2

Vesc = SQUARE ROOT 2 G 2M÷ R÷2 (2 OF R÷2 WILL BE MOVE UPWARD AND WILL MULTIPLY WITH 2 OF MASS )

Vesc' = SQUARE ROOT 2 G 4M÷R (square root 4 is equal to 2 so it will come out of square root )

Vesc' = 2 (SQUARE ROOT 2 G M÷R )

WE KNOW THAT SQUARE ROOT 2GM÷R IS EQUAL TO vesc

SO

Vesc' = 2 Vesc

Vesc' = 11.2 × 2 =22.4 km/sec

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Answer 2

Answer:

5.6 km/s

Explanation:

It was asked in jipmer 2015....

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