What is the acceleration due to gravity on the surface of a planet that has twice the mass of the earth and half its radius?

Hint: In order to solve this question, we should know that acceleration due to gravity on any planet is due to its mass and radius and here we will first note the acceleration due to gravity on earth and then rearranging the parameters value as given in the question and then will find the acceleration due to gravity on given surface of a planet.

Formula Used: The acceleration due to gravity on any planet is calculated as
${g_{planet}} = \dfrac{{G{M_{planet}}}}{{{R^2}_{planet}}}$ where,
g denotes acceleration due to gravity on a planet.
G is gravitational constant.
M is the mass of the planet.
R is the radius of the planet.


Complete step by step answer:
According to the question, we have given that
${M_{planet}} = 2{M_{earth}}$ mass of the planet is twice of mass of earth
${R_{planet}} = \dfrac{{{R_{earth}}}}{2}$ radius of the planet is half of the radius of earth
now, as we know that acceleration due to gravity of earth is ${g_{earth}} = 9.8m{s^{ - 2}}$ using formula we can write it as,
$\dfrac{{G{M_{earth}}}}{{{R^2}_{earth}}} = 9.8m{s^{ - 2}} \to (i)$
and acceleration due to gravity of planet can be written as
${g_{planet}} = \dfrac{{G{M_{planet}}}}{{{R^2}_{planet}}}$ on putting the value of mass and radius as ${M_{planet}} = 2{M_{earth}}$ ${R_{planet}} = \dfrac{{{R_{earth}}}}{2}$
we get,
${g_{planet}} = \dfrac{{8G{M_{earth}}}}{{{R^2}_{earth}}}$
and using equation (i) we get,
${g_{planet}} = 8 \times 9.8$
${g_{planet}} = 78.4m{s^{ - 2}}$
Hence, acceleration due to gravity on the planet is $78.4m{s^{ - 2}}.$

Note:It should be remembered that, acceleration due to gravity around an object is much is if its mass is large and its mainly due to general theory of relativity which says larger the mass of a body larger the bending of curvature of space-time which is the main reason of larger value of acceleration due to gravity.