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A spherical planet far out in space has mass ${{M}_{\circ }}$ and diameter ${{D}_{\circ }}$. A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to:

Last updated date: 12th Jul 2024
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Hint: From the above question it is very clear that we are asked to determine the acceleration due to gravity for an arbitrary planet with specific parameters. Acceleration due to gravity on any body having finite mass due to any arbitrary planet having specific mass ${{M}_{\circ }}$ and diameter ${{D}_{\circ }}$ is nothing but due to the gravitational force of attraction between the two bodies. Hence comparing Newton’s second law of motion with the universal law of gravitation between two bodies we will be able to determine the magnitude of acceleration due to gravity on a particle of mass m falling freely near the surface of this planet.

Complete step-by-step solution:
Let us say we have two bodies in space of masses M and m separated by a distance ‘r’. Hence from universal law of gravitation the force (F) between the two bodies is given by,
Where ‘G’ is the gravitational constant. The negative sign denotes that the force is attractive in nature.
Let us say we have a spherical planet of mass ${{M}_{\circ }}$and radius $\dfrac{{{D}_{\circ }}}{2}$. Let us say we have a particle of mass ‘m’ where $m<<{{M}_{\circ }}$ at a distance ‘d’ from the centre of the planet near the surface of the planet. Therefore from universal law of gravitation, we can imply that the force between the particle and the planet will be,
  & F=-\dfrac{Gm{{M}_{\circ }}}{{{\left( \dfrac{{{D}_{\circ }}}{2}+d \right)}^{2}}} \\
 & \because d<<<\dfrac{{{D}_{\circ }}}{2} \\
 & \therefore F=-\dfrac{Gm{{M}_{\circ }}}{{{\left( \dfrac{{{D}_{\circ }}}{2} \right)}^{2}}}.....(1) \\
From Newton’s second law we can imply that the force on a body is product of mass ‘m’ into acceleration ‘a’. Mathematically this can be represented as,
The body of mass ‘m’ gets accelerated towards the planet. Therefore comparing equation 1 and 2 we get,
  & a=\dfrac{G{{M}_{\circ }}}{{{\left( \dfrac{{{D}_{\circ }}}{2} \right)}^{2}}} \\
 & \Rightarrow a=\dfrac{G{{M}_{\circ }}}{\dfrac{{{D}_{\circ }}^{2}}{4}} \\
 & \therefore a=\dfrac{4G{{M}_{\circ }}}{{{D}_{\circ }}^{2}} \\
Therefore we can conclude that any body independent of its mass, will experience an acceleration due to gravity near the surface of the planet which is equal to $\dfrac{4G{{M}_{\circ }}}{{{D}_{\circ }}^{2}}$

Note: It is to be noted that the radius is half of the diameter of the sphere. The universal law of gravitation states that any two bodies in space having finite mass will experience a force which is proportional to the product of their respective masses and inversely proportional to the square of the distance between them. Force is mutual in nature hence both the bodies experience this force of attraction. When we consider planets with respect to bodies in free fall, the mass of the planets is huge with respect to heavenly bodies and hence the force is not sufficient enough to accelerate the planets.