Question

An imaginary planet has a mass $5$ times and radius $3$ times that of earth. What is the acceleration due to gravity on that planet if the acceleration due to earth is $10\,m{s^{ - 2}}$ ?

Answer 1

Hint:The formula of acceleration due to gravity states that the gravitational constant is multiplied with the mass of the body divided by the square of radius of that body. Apply this formula for both the body and by comparing them we can calculate the acceleration due to gravity on that planet.

Formula used:
The formula for acceleration due to gravity:
$g = \dfrac{{GM}}{{{r^2}}}$
Where, $r$= Radius of the body or planet, $G$= Gravitational Constant and $M$= Mass of the body or planet.

Complete step by step answer:
As per the problem given there is an imaginary planet whose mass is $5$ times the mass of earth and radius is $3$ times that of the earth. We know the acceleration due to gravity on earth is $10\,m{s^{ - 1}}$. The formula for acceleration due to gravity is as follow
$g = \dfrac{{GM}}{{{r^2}}}$

From the above formula we clearly see that the acceleration due to gravity is directly proportional to the mass of the planet and inversely proportional to the square of the radius of the planet. And here $G$ will remain the same for all types of body as it has a constant value $6.67 \times {10^{ - 11}}\,N\,K{g^{ - 2}}{m^2}$. Let g’ be the acceleration due to gravity on that imaginary planet, M’ be that mass of that planet and r’ be the radius of that planet.Now we can write
$g' = \dfrac{{GM'}}{{r{'^2}}}$
Here G is same as that of the earth

According to the acceleration due to gravity formula, when mass increases then the acceleration due to gravity increases hence if the mass increase $5$ times then the acceleration of gravity increase $5$ times and at the same time radius increases then the acceleration due to gravity decreases to the square of radius increases hence if the radius increase $3$ times then the acceleration of gravity decreases $9$ times.Now we can write
$g' = \dfrac{5}{9}g$
We know g is the Earth’s acceleration due to gravity which is $10m{s^{ - 2}}$ .
By putting the g value in the above equation we get,
$g' = \dfrac{5}{9} \times 10\,m{s^{ - 2}}$
$ \therefore g' = \dfrac{{50}}{9}m{s^{ - 2}}$

Hence acceleration due to gravity on that planet is $\dfrac{{50}}{9}m{s^{ - 2}}$.

Note:You can also solve it by putting the mass of the planet equals $5$ times the mass of earth and radius of the planet be $3$ times the mass of earth. Then put this changed mass and radius value in the place of the acceleration due to gravity on the imaginary planet formula.