Question

If mass of a planet is 3 times the mass of Earth and its radius is double the radius of earth find value of acceleration due to gravity on the surface of the planet ( value of g on Earth is 9.8m/s²)

Answer 1

Explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}\mid}}}$

Let's suppose name of planet be A

$\huge\bold{GIVEN:-} \\ \\ M_A = 3 \times M_E \\ \\ R_A = 2 \times R_E \\ \\ \large\purple{Formula:-} \\ \\ g = \frac{G M}{{(R) }^{2}} \\ \\ \\ \\ For \: \; Earth, \\ \\ \\ \\ g_E = \frac{G M_E}{{(R_E) }^{2}} = 9.8 m{s}^{-2}......(1) \\ \\ \\ \\ For \: \: Planet \: \: A, \\ \\ \\ \\ g_A = \frac{G M_A}{{(R_A) }^{2}} \\ \\ g_A = \frac{G \times 3 M_E}{{(2R_E) }^{2}} \\ \\ g_A = \frac{3}{4} \times \frac{G M_E}{{(R_E})^{2}}......(2) \\ \\ From \: \: (1) \: \: and \: \: (2) \\ \\ g_A = \frac{3}{4} \times 9.8 \\ \\ \huge\boxed{ g_A = 7.35 m{s}^{-2} }$

Answer 2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Let's say that planet to be planet A. Now acceleration due to gravity is represented by

g = \frac{GM}{ {r}^{2} } .

For earth let's say keywords to be :

• Mass of earth = $M_E$
• Radius of earth = $R_E$

Now, fir planet A this will be :-

• Mass of planet = $3M_E$
• Radius of planet = $2R_E$

We know that Acceleration due to gravity on earth is

$g = \frac{GM_e}{ {r_e}^{2} } = 9.8$

Now we know that acceleration due to gravity on Planet A will be

$g_a = \frac{G3M_e}{ {2r_e}^{2} }$

$g_a = \frac{3}{4} \times \frac{GM_e}{ {r_e}^{2} }$

Now after inputting the value we get :

$g_a = \frac{3}{4} \times 9.8 = \frac{29.4}{4}$

$g_a = \cancel \frac{29.4}{4} = 7.35 \frac{m}{ {s}^{2} }$

Therefore acceleration due to gravity on Planet A is $7.35m/ {s}^{2}$.

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BY SCIVIBHANSHU

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