Question

A planet is having a mass twice that of earth’s mass and its radius is 4 times to that of earth’s radius. Determine four times the acceleration due to gravity at the surface of this planet. Acceleration due to gravity at earth’s surface is $10m{s^{ - 2}}$.

Answer 1

Hint:Use Newton's law of gravitation to calculate the force on the test particle of mass m due to the planet and then divide that force with the m to get the acceleration due to gravity.

Complete step by step answer:
Now according to Newton's law of gravitation, every object attracts another object within its vicinity according to the given formula. The force of attraction depends upon the mass of the objects and the distance between them. Both the objects attract each other with equal amounts of force irrespective of their masses but their accelerations are inversely proportional to their masses. the interaction of smaller objects with massive planets causes the smaller object to accelerate towards the planet because it’s acceleration is much larger than the planet’s acceleration, and if the smaller objects mass is much less compared to the mass of the planet then this acceleration of the smaller object is approximately constant and it's called the acceleration due to gravity off the planet.
To calculate the gravitational acceleration, we must first calculate the gravitational force acting on a test particle due to the planet. Then we can divide the mass of the test particle (say m) from the gravitational force which will give us the gravitational acceleration of that planet.The gravitational force acting on a test particle is given by Newton's gravitational force law which is
$F = G\dfrac{{m{M_{planet}}}}{{{R_{planet}}^2}}$
${R_{planet}}$ is the radius of the planet
${M_{planet}} = 2{M_{earth}}$
${R_{planet}} = 4{R_{earth}}$
$\therefore F = G\dfrac{{m2{M_{earth}}}}{{{{\left( {4{R_{earth}}} \right)}^2}}}$
$F = \dfrac{1}{8}G\dfrac{{m{M_{earth}}}}{{{R_{earth}}^2}}$
Now that we have got the gravitational force acting on a test particle of mass m, we can calculate the acceleration of the particle due to the planet’s gravitational attraction
$g' = \dfrac{1}{8}G\dfrac{{{M_{earth}}}}{{{R_{earth}}^2}}$
${g_{earth}} = G\dfrac{{{M_{earth}}}}{{{R_{earth}}^2}} = 10m{s^{ - 2}}$
$\therefore g' = \dfrac{{{g_{earth}}}}{8} = \dfrac{{10}}{8} = 1.25m{s^{ - 2}}$
This is the value of the acceleration due to gravity at the surface of this planet therefore the answer would be four times this value which is $5m{s^{ - 2}}$.

Note:Another way to solve this question is to know the fact that acceleration due to gravity of a planet
$g' = \dfrac{{G{M_{planet}}}}{{{R_{planet}}^2}}$
$\because {M_{planet}} = 2{M_{earth}},{R_{planet}} = 4{R_{earth}}$
We could directly use this formula to calculate the acceleration due to gravity of the planet.