Question

A spherical planet has a mass $M_p$ and diameter $D_p$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to

• A. $\frac {4GM_p}{D_p^2} $
• B. $\frac {GM_pm}{D_p^2} $
• C. $\frac {GM_p}{D_p} $
• D. $\frac {4GM_pm}{D_p^2} $

Answer 1

The Correct Option is A

To solve this problem, we need to determine the acceleration due to gravity experienced by a particle near the surface of a spherical planet. This problem utilizes the fundamental formula for gravitational force and acceleration due to gravity. Let's break down the solution step by step:

1. Gravitational Force: The force of gravity acting on a particle of mass $m$ near the surface of a planet is given by Newton's law of universal gravitation: $$ F = \frac{G M_p m}{r^2} $$ where $G$ is the gravitational constant, $M_p$ is the mass of the planet, and $r$ is the radius of the planet.
2. Relation to Acceleration: Acceleration due to gravity, $g$, is the force per unit mass. Thus, it is calculated as: $$ g = \frac{F}{m} = \frac{G M_p}{r^2} $$ This expresses the acceleration experienced by the mass $m$.
3. Expressing in Terms of Diameter: Since the problem specifies the diameter $D_p$ of the planet, we need to express $r$ in terms of $D_p$. The radius $r$ is half of the diameter: $$ r = \frac{D_p}{2} $$
4. Substitute Radius in Gravity Formula: Substituting $r = \frac{D_p}{2}$ into the expression for $g$, we have: $$ g = \frac{G M_p}{\left(\frac{D_p}{2}\right)^2} = \frac{G M_p}{\frac{D_p^2}{4}} $$ Simplifying this, we find: $$ g = \frac{4 G M_p}{D_p^2} $$
5. Conclusion: Hence, the particle experiences an acceleration due to gravity equal to $\frac {4GM_p}{D_p^2}$. This matches the given correct answer. The other options involving mass $m$ or wrong dimensions do not represent the formula for gravitational acceleration.