Answer:
[tex]33.25 m/s^2[/tex]
Explanation:
The strength of the gravitational field at a certain distance [tex]r[/tex] from the centre of a planet is given by:
[tex]g=\frac{GM}{r^2}[/tex]
where:
G is the gravitational constant
M is the mass of the planet
The equation can be rewritten as follows:
[tex]gr^2 = GM[/tex]
The term on the right is a constant term, so we can write:
[tex]g_1 r_1^2 = g_2 r_2^2[/tex]
where
[tex]g_1[/tex] is the gravitational field strength at distance [tex]r_1[/tex]
[tex]g_2[/tex] is the gravitational field strength at distance [tex]r_2[/tex]
Here we have:
When [tex]r_1 = 5R[/tex] (the distance is 5 planetary radii), the gravitational field strength is
[tex]g_1 = 1.33 m/s^2[/tex]
where R is the radius of the planet.
Instead at the surface
[tex]r_2 =R[/tex]
And so the strength of the gravitational field at the surface is:
[tex]g_2 = \frac{g_1 r_1^2}{r_2^2}=\frac{(1.33)(5R)^2}{R^2}=(1.33)(25)=33.25 m/s^2[/tex]
