Question

If the mass of the earth is $80$ times that of a planet and diameter is double that of a planet and '$\mathrm{g}$' on the earth is $9.8{\mathrm{ms}}^{-2}$ , then the value of '$\mathrm{g}$' on that planet is = ______ ${\mathrm{ms}}^{-2}$

• A. $4.9$
• B. $0.98$
• C. $0.49$
• D. $49$

Answer 1

The correct option is C

$0.49$

Explanation for correct answer:

Option (C) correct answer:

Step 1 Given data:

The mass of the earth(${\mathrm{M}}_{\mathrm{e}}$) is $80$ times that of a planet(${\mathrm{M}}_{\mathrm{p}}$).

That is, ${\mathrm{M}}_{\mathrm{e}}=80{\mathrm{M}}_{\mathrm{p}}$

The diameter of the earth is double that of a planet, so radius of the earth(${\mathrm{R}}_{\mathrm{e}}$) is also double that of the radius of a planet(${\mathrm{R}}_{\mathrm{p}}$).

That is, ${\mathrm{R}}_{\mathrm{e}}=2{\mathrm{R}}_{\mathrm{p}}$

The acceleration due to gravity value on earth is, $\mathrm{g}=9.8{\mathrm{ms}}^{-2}$

Step 2 Formula used:

The acceleration due to gravity is, $\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$

Where,

$\mathrm{G}$ is the gravitational constant

$\mathrm{M}$ is the Mass

$\mathrm{R}$ is the radius

Step 3 Value of '$\mathrm{g}$' on that planet:

Let the acceleration due to gravity of planet is ${\mathrm{g}}_{\mathrm{e}}=\frac{{\mathrm{GM}}_{\mathrm{e}}}{{{\mathrm{R}}_{\mathrm{e}}}^2}....\left(1\right)$

Let the acceleration due to gravity of planet is ${\mathrm{g}}_{\mathrm{p}}=\frac{{\mathrm{GM}}_{\mathrm{p}}}{{{\mathrm{R}}_{\mathrm{p}}}^2}....\left(2\right)$

Divide equation (2) by equation (1),

$$\begin{gathered} \frac{{\mathrm{g}}_{\mathrm{p}}}{{\mathrm{g}}_{\mathrm{e}}}=\frac{\frac{{\mathrm{GM}}_{\mathrm{p}}}{{{\mathrm{R}}_{\mathrm{p}}}^2}}{\frac{{\mathrm{GM}}_{\mathrm{e}}}{{{\mathrm{R}}_{\mathrm{e}}}^2}} \\ \frac{{\mathrm{g}}_{\mathrm{p}}}{{\mathrm{g}}_{\mathrm{e}}}=\frac{{\mathrm{GM}}_{\mathrm{p}}}{{{\mathrm{R}}_{\mathrm{p}}}^2}\times \frac{{{\mathrm{R}}_{\mathrm{e}}}^2}{{\mathrm{GM}}_{\mathrm{e}}} \\ \frac{{\mathrm{g}}_{\mathrm{p}}}{{\mathrm{g}}_{\mathrm{e}}}=\frac{{\mathrm{M}}_{\mathrm{p}}\times {{\mathrm{R}}_{\mathrm{e}}}^2}{{{\mathrm{R}}_{\mathrm{p}}}^2\times {\mathrm{M}}_{\mathrm{e}}}.....\left(5\right) \end{gathered}$$

It is given that,

${\mathrm{M}}_{\mathrm{e}}=80{\mathrm{M}}_{\mathrm{p}}.....\left(3\right)$

${\mathrm{R}}_{\mathrm{e}}=2{\mathrm{R}}_{\mathrm{p}}....\left(4\right)$

Substitute these equation (3) and (4) in equation (5),

$$\begin{gathered} \frac{{\mathrm{g}}_{\mathrm{p}}}{{\mathrm{g}}_{\mathrm{e}}}=\frac{{\mathrm{M}}_{\mathrm{p}}\times {\left(2{\mathrm{R}}_{\mathrm{p}}\right)}^2}{{{\mathrm{R}}_{\mathrm{p}}}^2\times 80{\mathrm{M}}_{\mathrm{p}}} \\ \frac{{\mathrm{g}}_{\mathrm{p}}}{{\mathrm{g}}_{\mathrm{e}}}=\frac{{\mathrm{M}}_{\mathrm{p}}\times 4{{\mathrm{R}}_{\mathrm{p}}}^2}{{{\mathrm{R}}_{\mathrm{p}}}^2\times 80{\mathrm{M}}_{\mathrm{p}}} \\ \frac{{\mathrm{g}}_{\mathrm{p}}}{{\mathrm{g}}_{\mathrm{e}}}=\frac{4}{80}.....\left(6\right) \end{gathered}$$

It is given that ${\mathrm{g}}_{\mathrm{e}}=9.8{\mathrm{ms}}^{-2}$

Substituting this in equation (6),

$$\begin{gathered} \frac{{\mathrm{g}}_{\mathrm{p}}}{9.8}=\frac{4}{80} \\ {\mathrm{g}}_{\mathrm{p}}=\frac{4}{80}\times 9.8 \\ {\mathrm{g}}_{\mathrm{p}}=0.49{\mathrm{ms}}^{-2} \end{gathered}$$

Therefore, the value of '$\mathrm{g}$' on that planet is $0.49{\mathrm{ms}}^{-2}$