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Which of the following is the acceleration due to gravity on the surface of the sun ?(Given gravitation constant G is $6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ and the radius of the sun is $6.96 \times 10^{8} m$ , mass of sun is $1.99 \times 10^{30} kg$ )

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274 m s^{- 2}

2.74 m s^{- 2}

27.4 m s^{- 2}

0.274 m s^{- 2}

answer is A.

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Detailed Solution

Acceleration due to gravity on the surface of the sun is

274

m s^{- 2}

.
As per Newton's universal law of gravitation we know that force of attraction between 2 bodies is given by

F = \frac{G m_{1} m_{2}}{r^{2}} - - - (i)

where

G

is the gravitational constant

m_{1}

is the mass of first body

m_{2}

is the mass of the second body
r is the distance between two bodies.
At the same time we know that

F = mg - - - (ii)

From (i) and (ii) we can write

\frac{G {m_{2} m}_{1}}{r^{2}} = m_{2} g

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

Mass of the sun

M = 1.99 \times 10^{30} kg

On the surface of the sun, the distance to the center of mass is same as the radius of the sun

R = 6.96 \times 10^{8} m

m

is the mass of the earth

r

is the radius of the earth

g_{e}

is the acceleration due to gravity on the surface of the earth

g_{s}

is the acceleration due to gravity on the surface on the sun
Here we need to find the acceleration due to gravity on the surface of the sun.
Hence we can write from equation (i)

\frac{G m_{1} m_{2}}{r^{2}} = \frac{GMm}{R^{2}} - - - (a)

Also from (ii)

F = m g_{s} - - - (b)

Equating (a) and (b)

\frac{GMm}{R^{2}} = m g_{s}

\Rightarrow \frac{GM}{R^{2}} = g_{s} - - - (c)

Substituting the values in (c)
We get

{\Rightarrow g}_{s} = \frac{6.67 \times 10^{- 11} \times 1.99 \times 10^{30}}{{(6.96 \times 10^{8})}^{2}}

{\Rightarrow g}_{s} = \frac{0.274 \times 10^{- 11} \times 10^{30}}{{{(10}^{8})}^{2}}

{\Rightarrow g}_{s} = 0.274 \times 10^{- 11} \times 10^{30} \times 10^{- 16}

{\Rightarrow g}_{s} = 0.274 \times 10^{3}

{\Rightarrow g}_{s} = 274 m s^{- 2}

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