Question

Find the relation between the gravitational field on the surface of two planets A and B of masses ${{m}_{A}}$​, and ${{m}_{B}}$ and radii ${{R}_{A}}$ and ${{R}_{B}}$ respectively, if
a. they have equal mass,
b. they have equal (uniform) density.

A: a) for ${{m}_{A}}={{m}_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{B}}^{2}}{{{R}_{A}}^{2}}$
b) for ${{\rho }_{A}}={{\rho }_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{A}}^{{}}}{3{{R}_{B}}^{{}}}$

B: a) for ${{m}_{A}}={{m}_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{B}}^{2}}{{{R}_{A}}^{2}}$
b) for ${{\rho }_{A}}={{\rho }_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{A}}^{{}}}{{{R}_{B}}^{{}}}$

C: a) for ${{m}_{A}}={{m}_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{B}}^{2}}{{{R}_{A}}^{2}}$
b) for ${{\rho }_{A}}={{\rho }_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{2{{R}_{A}}^{{}}}{{{R}_{B}}^{{}}}$

D: a) for ${{m}_{A}}={{m}_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{B}}^{2}}{{{R}_{A}}^{2}}$
b) for ${{\rho }_{A}}={{\rho }_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{2{{R}_{A}}^{{}}}{2{{R}_{B}}^{{}}}$

Answer 1

To find the relationship between gravitational fields, we have to try and relate their gravitational field intensities with respect to the mass as well as the densities. We can relate mass and further express it in terms of density and equate the relations to solve this problem.

Formula used:
Formula for gravitational field intensity:
$E=\dfrac{GM}{{{R}^{2}}}$, where E is the gravitational field intensity, G is the gravitational constant, M is the mass of the body and R is the radius or the distance of separation.

Complete step by step solution:
We are given the detail regarding two planets A and B. Let the gravitational field intensity on the surfaces of A and B be ${{E}_{A}}$ and ${{E}_{B}}$ respectively.
We know that their masses are ${{m}_{A}}$​, and ${{m}_{B}}$ and radii are ${{R}_{A}}$ and ${{R}_{B}}$ respectively.
$E=\dfrac{GM}{{{R}^{2}}}$
Hence we can write ${{E}_{A}}$ and ${{E}_{B}}$ as
${{E}_{A}}=\dfrac{G{{M}_{A}}}{{{R}_{A}}^{2}}=\dfrac{G\dfrac{4}{3}\pi {{\rho }_{A}}{{R}_{A}}}{{{R}_{A}}^{2}} \\\ {{E}_{B}}=\dfrac{G{{M}_{B}}}{{{R}_{B}}^{2}}=\dfrac{G\dfrac{4}{3}\pi{{\rho }_{B}}{{R}_{B}}}{{{R}_{B}}^{2}} \\\ \\\$ (since mass is the product of density and volume)
Upon equating the above two equations, we can arrive at the conclusion that
a) for ${{m}_{A}}={{m}_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{B}}^{2}}{{{R}_{A}}^{2}}$ and
b) for ${{\rho }_{A}}={{\rho }_{B}},\dfrac{{{E}_{A}}}{{{E}_{B}}}=\dfrac{{{R}_{A}}^{{}}}{{{R}_{B}}^{{}}}$

So, the correct answer is “Option B”.

Note:
Gravitational field intensity states that if we bring a unit mass from infinity to a gravitational field, then a gravitational force acts on it due to a comparatively bigger mass for which the field is created . This force is the gravitational field intensity.