Question

The mass of a planet is 6$\times$10²⁴ kg and its diameter is 12.8 $\times$10³ km. If the value of a gravitational constant is 6.7 $\times$10^$-$11 Nm²/kg², calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be?

Answer 1

The value of acceleration due to gravity is given by the relation, $g=\frac{GM}{R^2}$ (Where 'M' is the mass and 'R' is the radius of the planet)

Mass of the planet = 6$\times$10²⁴ kg

The radius of the planet = $\frac{\mathrm{Diameter}}{2}=\frac{12.8\times {10}^3\mathrm{km}}{2}=6.4\times {10}^3\mathrm{km}$

Gravitational constant = 6.7 $\times$10^$-$11 Nm²/kg²

So, $g=\frac{6.7\times {10}^{-11}{\mathrm{Nm}}^2{\mathrm{kg}}^{-2}\times 6\times {10}^{24}\mathrm{kg}}{(6.4\times {10}^3\times {10}^3\mathrm{m}{)}^2}=9.8{\mathrm{ms}}^{-2}$

This planet could be earth because the value of acceleration due to gravity is the same on the planet earth.