Question

Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its surface, in terms of the free-fall acceleration g at the surface of the earth?

Answer 1

Answer:

2 g/9

Explanation:

mass of planet, Mp = 2 x Me

radius of planet, Rp = 3 x Re

Where, Me is the mass of earth and Re is the radius of earth.

The formula for acceleration due to gravity on earth is given by

$g = \frac{GM_{e}}{R_{e}^{2}}$     .... (1)

The acceleration due to gravity on the planet is given by

$g' = \frac{GM_{p}}{R_{p}^{2}}$

By substituting the values, we get

$g' = \frac{2GM_{e}}{9R_{e}^{2}}$    ..... (2)

Divide equation (2) by equation (1), we get

g'/g = 2/9

g' = 2 g/9

Thus, the acceleration due to gravity on th enew planet is 2 g/9.  

Answer 2

Answer:

$g_n=\dfrac{2}{9}g$

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

$g=\dfrac{GM}{R^2}$

On new planet

$g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}$

Dividing the two equations we get

$\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g$

The acceleration due to gravity on the other planet is $g_n=\dfrac{2}{9}g$