Question

Which of the following are correct?
This question has multiple correct options.
A. If $R$ is the radius of a planet, $g$ is the acceleration due to gravity, the mean density of the planet is $3g/4\pi GR$ .
B. Acceleration due to gravity is a universal constant.
C. The escape velocity of a body from earth is $11.2\,{\text{km}}\,{{\text{s}}^{ - 1}}$ . The escape velocity from a planet which has double the mass of earth and half its radius is $22.4\,{\text{km}}\,{{\text{s}}^{ - 1}}$ .
D. The ratio of gravitational mass and inertial mass of a body at the surface of earth is $1$ .

Answer 1

First of all, we will find the expression of $g$ and then expand it to bring the term density into it. We will manipulate accordingly and find the density. We can jump more on the moon’s surface due to less gravity.

Complete step by step answer:
In the given question, we are supplied the following data:
There are four options, which may be correct or wrong.
In the radius of the planet is $R$ , $g$ is the acceleration due to gravity.
We are to find the mean density of the planet.
We can write as:
g = \dfrac{{G \times M}}{{{R^2}}} \\\ \Rightarrow g = \dfrac{{G \times V \times \rho }}{{{R^2}}} \\\ \Rightarrow g = \dfrac{G}{{{R^2}}} \times \dfrac{4}{3}\pi {R^3}\rho \\\ \Rightarrow \rho = \dfrac{{3g}}{{4\pi GR}} \\\
Hence, the mean density of the planet is $\dfrac{{3g}}{{4\pi GR}}$ .

We know, acceleration due to gravity is the acceleration at which an object will fall down when released from a height. The value of $g$ is not constant, it increases as the distance from it increases. Value of $g$ is lowest at great heights and highest at small heights. The gravity at equator is $9.78\,{\text{m}}\,{{\text{s}}^{ - 2}}$ while the gravity at poles is $9.83\,{\text{m}}\,{{\text{s}}^{ - 2}}$ .

We know that the escape velocity is given by the expression:
${v_{\text{e}}} = \sqrt {\dfrac{{2GM}}{R}}$
But if the mass is doubled and radius is halved. Then the escape velocity will become:
{v_{\text{E}}} = \sqrt {\dfrac{{2GM}}{R}} \\\ \Rightarrow {v_{\text{E}}} = \sqrt {\dfrac{{2G\left( {2 \times M} \right)}}{{\dfrac{R}{2}}}} \\\ \Rightarrow {v_{\text{E}}} = 2 \times \sqrt {\dfrac{{2GM}}{R}} \\\ \Rightarrow {v_{\text{E}}} = 2 \times 11.2\,{\text{km}}\,{{\text{s}}^{ - 2}} \\\
$\therefore {v_{\text{E}}} = 22.4\,{\text{km}}\,{{\text{s}}^{ - 2}}$
Hence, the escape velocity becomes $22.4\,{\text{km}}\,{{\text{s}}^{ - 1}}$ . According to the theory of general relativity which uses the equivalence principle which implies that the two masses are identical. Hence the ratio of inertial mass to gravitational mass is equal to $1$ .

The correct options are A, B and D.

Note: While solving this problem, most of the students seem to have a confusion regarding the acceleration due to gravity. They make a misunderstanding that it is highest at the equator. However, this is wrong. We know, acceleration due to gravity is inversely proportional to the square of the distance. It is important to remember that higher the gravity of a planet, it is tougher to escape its velocity.