Question

What is the acceleration due to gravity on the surface of a planet that has twice the mass of the Earth and half its radius?

Answer 1

The mass increases linearly but radius decreases exponentially, so the result is
`9.8 m/s^2 * 2 * 4 = 78.4 m/s^2`

Explanation:

Let's first look at the equation for the force of gravity:

`F_g=G(m_1m_2)/r^2`

which is often simplified for working with objects on the surface of the Earth (since we know the gravitational constant and the mass of the Earth) to

`F_g=M/r^2`

where M is the mass experiencing Earth's gravity.

So what happens when we double the mass of the Earth and reduce its radius to 1/2? Let's multiply `m_1` by 2 and substitute in `1/2 r` for `r`. So first start with the full equation:

`F_g=G(m_1m_2)/r^2`

then make the substitutions:

`F_g=G((2m_1)m_2)/(1/2r)^2`

`F_g=G((2m_1)m_2)/(1/4r)`

So the numerator increases linearly (`xx2` ) but the denominator reduces by an exponential - in this case (`xx4` ).

The force of gravity on Earth is roughly `9.8 m/s^2` but on this other planet, it would be:

`9.8 m/s^2 * 2 * 4 = 78.4 m/s^2`