Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then :

Question

Two identical particles each of mass \( m \) go around a circle of radius \( a \) under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

To determine the acceleration due to gravity on the surface of a new planet, we begin by understanding the relationship between density, volume, and the gravitational force.

Let's recall the formula for gravitational acceleration on the surface of a planet:

g = \frac{GM}{R^2}

where:

G is the universal gravitational constant.
M is the mass of the planet.
R is the radius of the planet.

Given that the new planet has the same density as Earth and is 3 times larger in size:

The radius of the new planet, R', is 3 times the radius of Earth, R:

R' = 3R

Since the densities of the earth and the new planet are the same, we have:

\frac{M}{V} = \frac{M'}{V'}

The volume of a sphere is given by V = \frac{4}{3}\pi R^3.

Thus, for our new planet:

V' = \frac{4}{3}\pi (3R)^3 = \frac{4}{3}\pi 27R^3 = 27V

This implies:

M' = 27M

Now substituting back into the formula for gravity, the gravitational acceleration on the new planet is:

g' = \frac{GM'}{(R')^2} = \frac{G(27M)}{(3R)^2} = \frac{27GM}{9R^2} = 3 \left(\frac{GM}{R^2}\right)

Therefore:

g' = 3g

Thus, the correct answer is g' = 3g. This proves that the gravitational acceleration on the surface of the new planet is three times that on the surface of the Earth.

Answer 2