Question

A newly discovered exoplanet has triple the mass and double the radius as compared to Earth. What is the approximate surface gravity on this exoplanet?​

Answer 1

Given:

A newly discovered exoplanet has triple the mass and double the radius as compared to Earth.

To find:

Approximate surface gravity on this new planet.

Calculation:

General expression for gravitational acceleration on any planet of mass m and radius r is given as follows:

$\boxed{ \sf{g_{p} = \dfrac{Gm}{ {r}^{2} } }}$

So, for earth:

$\therefore \: \sf{g_{e} = \dfrac{Gm_{e} }{ ({r_{e}) }^{2} } }$

So, for the planet:

$\therefore \: \sf{g_{p} = \dfrac{Gm_{p} }{ ({r_{p}) }^{2} } }$

Dividing the equations:

$\sf{ \therefore \: \dfrac{g_{p}}{g_{e}} = \dfrac{m_{p}}{m_{e}} \times \dfrac{ {(r_{e})}^{2} }{ {(r_{p})}^{2} } }$

$\sf{ = > \: \dfrac{g_{p}}{g_{e}} = \dfrac{3m_{e}}{m_{e}} \times \dfrac{ {(r_{e})}^{2} }{ {(2r_{e})}^{2} } }$

$\sf{ = > \: \dfrac{g_{p}}{g_{e}} = \dfrac{3}{4} }$

$\sf{ = > \: g_{p} = g_{e} \times \dfrac{3}{4} }$

$\sf{ = > \: g_{p} = 10\times \dfrac{3}{4} }$

$\boxed{ \sf{ = > \: g_{p} = 7.5 \: m {s}^{ - 2} }}$

HOPE IT HELPS.