Question

A spacecraft needs to orbit planet x at a distance from the surface of 14,000 km. if the radius of the planet is 7000 km and the orbital period is 15 hours, what is the acceleration due to gravity on the surface of the planet?

Answer 1

Let's do it algebraically by first defining some variables. Use SI units.
Radius of the planet: $r_{0}$
Distance between the spacecraft and the center of the planet: $r_{1} = r_{0} + 14000000$
Mass of the planet: $M$
Mass of the spacecraft: $m$
Tangential velocity of the spacecraft in its orbit: $v$
Angular velocity of the spacecraft: $w= \frac{v}{ r_{1} }$
Orbital period of the spacecraft: $T = \frac{2 \pi }{w}$

Since the centripetal force acting on the spacecraft is the gravitational force from the planet at distance $r_{1}$ from the planet's centre:
$F_{c} = F_{g} ;\\ m \frac{v^{2}}{ r_{1} } = G \frac{mM}{r_{1}^{2} }$
Note we can cancel out the mass of the spacecraft.
Using the formula in the definitions above, substitute:
$\frac{v^{2} }{r_{1} } = w^2 r_{1} = \frac{4 \pi^2 }{T^2} r_{1}$
So we can substitute for this expression and obtain
$\frac{4 \pi ^2}{T^2} r_{1} =G \frac{M}{ r_{1}^2 } ; \\ M = \frac{4 \pi ^2 r_{1}^3 }{T^2 G}$
So now that we have the mass of the planet, we can easily calculate that $g$, the acceleration due to gravity on its surface, is given by:
$g=G \frac{M}{ r_{0}^2 } = \frac{G}{ r_{0}^2 } \frac{4 \pi ^2 r_{1}^3 }{T^2G} = \frac{4 \pi ^2 r_{1}^3 }{ r_{0}^2 T^2 }$