Question

The radius of the Earth is reduced by
4% of its intial value .The mass
of the earth remains unchanged.
What will be the percentage of
change in the escape velocity?​

Answer 1

Answer:

The acceleration due to gravity at the surface of earth is g = G m / R2 .

The radius of eath decreases by 4 %, therefore

R' = R - 0.04 R = 0.96R.

Now acceleration due to gravity will be :

g' = GM / ( 0.96 R )2 = 1.09 g

Percentage increase in acceleration due to gravity

[(g' - g )/g ] x 100 = 9 %

We hope this clarifies your doubt.

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Answer 2

Given that,

The radius of the Earth is reduced by  4% of its initial value .

The mass  of the earth remains unchanged.

We know that,

The acceleration due to gravity at the surface of earth is

$g=\dfrac{Gm}{R^2}$

The radius of the Earth is reduced by  4% of its initial value.

So, $R'=R-0.-04R$

$R'=0.96R$

We need to calculate the new acceleration due to gravity

Using formula of acceleration

$g'=\dfrac{GM}{R'^2}$

We need to calculate the percentage of  change in the escape velocity

Using formula of acceleration

$\dfrac{g'-g}{g}\times100=\dfrac{\dfrac{GM}{R'^2}-\dfrac{GM}{R^2}}{\dfrac{GM}{R^2}}\times100$

Put the value into the formula

$\dfrac{g'-g}{g}\times100=\dfrac{\dfrac{GM}{(0.96R)^2}-\dfrac{GM}{R^2}}{\dfrac{GM}{R^2}}\times100$

$\dfrac{g'-g}{g}\times100=(\dfrac{1}{(0.96)^2}-1)\times100$

$\dfrac{g'-g}{g}\times100=9\%$

Hence, The percentage of  change in the escape velocity is 9%.