Question

9. The mass of a planet is 10^18 kg and its radius is 10^3 m. Calculate the acceleration due to gravity on its surface. Use the known value
of G. (Ans. 66.7 m 54)​

Answer 1

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

Law of Universal gravitation says that the force of attraction between two bodies directly proportional to product of their masses and inversely proportional to square of distance between them. This law is also used to find acceleration due to gravity on an object. The expression based on this law and used to calculate acceleration due to gravity is :

$\maltese \: g = \frac{Gm}{ {r}^{2} }$

In this equation :

g = acceleration due to gravity

G = Universal gravitational constant

m = mass of body

r = it's radius

$\maltese \: \blue{\mid{\underline{\overline{\textbf{Coming to question :-}}}\mid}}$

In this question we have been given the mass of a planet and it's radius. Acceleration due to gravity of this planet is not known. We have to calculate it's acceleration due to gravity.

It is given that :-

Mass of planet = 10^18kg

Radius of planet = 1000m

G = 6.673 × 10^-11

Acceleration due to gravity = g = ?

Now after inputting the known values in the equation we get :

$\mapsto \: g = \frac{6.673 \times {10}^{ - 11} \times {10}^{18} }{ {1000}^{2} }$

After solving a bit we get :

$\mapsto \: g = 66.7m \frac{m}{ {s}^{2} }$

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BY SCIVIBHANSHU

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Answer 2

Answer:

$66.7 \: m {s}^{ - 2}$

Explanation:

$mass \: of \: planet \: = {10}^{18} \\ radius \: of \: the \: planet = {10}^{3} m \\ value \: of \: g \: (capital) = \: 6.67 \times {10}^{ - 11} n {m}^{2} \: k {g}^{ - 2}$

$therefore \}$

$g \: = \frac{6.67 \times {10}^{ - 11} \times {10}^{18} }{( {10}^{3) {}^{2} } }$

$= \frac{6.67 \times {10}^{18}}{ {10}^{6} \times {10}^{11} }$

$= 6.67 \times 10$

$= 66.7m {s}^{ - 2}$