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The acceleration due to surface of the earth is $9.8 {ms}^{- 2}$ .If the radius of the orbit of the spaceship from the center of the earth is $2 R$ , where $R$ is the radius of the earth. Which of the following is the acceleration due to gravity at the spaceship? (Gravitational constant $G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ )

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2.45 m s^{- 2}

24.5 m s^{- 2}

0.245 m s^{- 2}

0.0245 m s^{- 2}

answer is A.

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Detailed Solution

The acceleration due to gravity on the spaceship is

2.45 m s^{- 2}

.
As per Newton’s law of gravitation we know that

F = \frac{G m_{1} m_{2}}{d^{2}} . . . . . . . . . (a)

where

G

is the gravitational constant (

6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

)

m_{1}

is the mass of the first object.

m_{2}

is the mass of the second object

d

is the distance between them
From (a) we can write

F = \frac{G m_{1} m_{2}}{R^{2}} . . . . . . . . . . (b)

where

G

is the gravitational constant

m_{1}

is the mass of the earth.

m_{2}

is the mass of the spaceship.

R

is the radius of earth.
At the same time

F = mg . . . . . . . . . . . (c)

where

F

is the force of attraction

m

is the mass of the body (Here it is

m_{2}

)

g

is the acceleration due to gravity.
We can equate equations (b) and (c ) So that we get

\frac{G m_{1} m_{2}}{R^{2}} = m_{2} g

\Rightarrow g = \frac{G m_{1}}{R^{2}} . . . . . . . . . . . . . . . (i)

Spaceship is at a distance of

2 R

from the center of the earth, therefore acceleration due to gravity of the spaceship become

\Rightarrow g^{'} = \frac{G m_{1}}{{(2 R)}^{2}}

\Rightarrow g^{'} = \frac{G m_{1}}{{4 R}^{2}} . . . . . . . . . . . . . . . . . (ii)

For finding the value of acceleration due to gravity on spaceship which is at a distance of

2 R

from the center of the earth we can divide equation

(i

i) with

(i)

.
So we get

\frac{g^{'}}{g} = \frac{\frac{G m_{1}}{{4 R}^{2}}}{\frac{G m_{1}}{R^{2}}}

\Rightarrow \frac{g^{'}}{g} = \frac{G m_{1}}{{4 R}^{2}} \times \frac{R^{2}}{G m_{1}}

\Rightarrow \frac{g^{'}}{g} = \frac{1}{4}

\Rightarrow g^{'} = \frac{g}{4}

{\Rightarrow g}^{'} = \frac{9.8}{4} m s^{- 2}

So the value of

g^{'} = 2.45 m s^{- 2}

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The acceleration due to surface of the earth is 9.8 ms-2.If the radius of the orbit of the spaceship from the center of the earth is 2R, where R is the radius of the earth. Which of the following is the acceleration due to gravity at the spaceship? (Gravitational constant G =6.67×10-11 N m 2kg-2)

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Detailed Solution

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