Question

Calculate the value of acceleration due to gravity on the surface of a planet, which has mass, twice as that of Earth, and radius, half as that of Earth.

Answer 1

Ans: 78.48 $\frac{m}{se{c}^2}$
Explanation:

As we know

(1) acceleration due to gravity on the surface of any planet can be given by $g_p=\frac{G{M}_p}{{R^2}_p}$ ;

where $M_p=Massoftheplanetand{R}_p=Radiusoftheplanet$

(2) acceleration due to gravity on the surface of earth can be given by $g_e=\frac{G{M}_e}{R_e^2}$ ;

where $M_e=Massoftheearthand{R}_e=Radiusoftheearth$

given ;

$M_P=2M_e$

$R_p=\frac{R_e}{2}$

solution :

$g_p=\frac{G{M}_p}{R_p^2}$

by substituting

$M_{p}and{R}_{p}valuesintermsof{M}_{e}and{R}_{e}respectively$in $g_p$

$g_p=\frac{G\left(2M_e\right)}{\frac{R_e^2}{4}}$ $\Rightarrow 8\frac{G{M}_e}{R_e^2}\Rightarrow 8g_e$ ; $\because g_e=\frac{G{M}_e}{R_e^2}$

and we also know that $g_e=9.81\frac{m}{se{c}^2}$

$g_p=8\left(9.81\right)\frac{m}{{sec}^2}$ $\Rightarrow 78.48\frac{m}{se{c}^2}$

Hence, the correct answer should be $g_p=78.48\frac{m}{se{c}^2}$