The mass of a planet is $6\times10^{24}\ kg$ and its diameter is $12.8\times 10^3\ km$. If the value of gravitational constant be $6.7\times10^{-11}\ Nm^2/kg^2$, calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be?

Given,

Maas of planet$=6\times10^{24}\ kg$

Diameter of the planet $=12.8\times10^3\ km=12.8\times10^6\ m$

Therefore, radius of planet $=\frac{1}{2}\times12.8\times10^6\ m=6.4\times10^6\ m$

$G=6.7\times10^{-11}\ Nm^2/kg^2$

As we know that,

Gravitational acceleration, $g=\frac{(Gm)}{r^2}$

Or $g=(6.7\times10^{-11})\times\frac{(6\times10^{24})}{(6.4\times10^6)^2}$

Or $g=9.81\ m/s^2$

On the earth value of gravitational acceleration, $g$ is $9.8\ m/s^2$, Therefore, this planet is the earth.
 

Updated on: 10-Oct-2022

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