Answer:
[tex]33.25 m/s^2[/tex]
Explanation:
The strength of the gravitational field produced by a planet at a distance r from the centre of the planet is given by
[tex]g=\frac{GM}{r^2}[/tex]
where
G is the gravitational constant
M is the mass of the planet
r is the distance from the centre of the planet
We can rewrite the equation as
[tex]gr^2 = GM[/tex]
And since the term on the right is constant, we can write
[tex]g_1 r_1^2 = g_2 r_2^2[/tex]
where [tex]g_1,g_2[/tex] are the strength of the gravitational field measured at two difference distances r1, r2.
In this problem we have:
[tex]g_1 = 1.33 m/s^2[/tex] when the distance is 5 planetary radii, so
[tex]r_1 = 5R[/tex]
where R is the planetary radius.
Therefore, at the surface,
[tex]r_2 =R[/tex]
And the strength of the gravitational field 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]