Question

A spacecraft is in orbit around a planet. The radius of the orbit is 2.9 times the radius of the planet (which is R = 71498 km). The gravitational field at the surface of the planet is 24 N/kg. What is the period of the spacecraft's orbit? [Hint: You don't need any more data about the planet to solve the problem.]

Answer 1

Radius of the orbit= 2.9 \(\times\) 71498= 207344.2 km
We have Gravitational Field at the surface as 24 N/KG
Let E be gravitational field
\[E= G\frac{M}{R^2}\]
G=universal gravitational constant
M= mass of planet, Let's find the mass
\[M= \frac{E\times R^2}{G}\]
We get mass of planet
\[M=\frac{24N/kg\times (71498*1000 m)^2}{6.673\times 10^{-11} N-m^2/kg^2}\]
we get
\[\large{M=1.83\times 10^{27} kg}\]
From Kepler's third law
time period of a satellite, T
\[T=\sqrt{\frac{4\times \pi\times r^3}{G\times M}}\]
r= radius of orbit
\[T=\sqrt{\frac{4\times\pi\times(207344.2 *1000m)^3}{6.673\times10^{-11} Nm^2/kg^2\times 1.83\times10^{27} kg }}\]
Now you can find the period, can't you??

Answer 2

a slight correctiom
\[ T=\sqrt{\frac{4\times\pi^2\times r^3}{G\times M}}\]
now substitute the values and find the orbital period

Answer 3

is that given in Seconds? Because It has to be in hours..and I need to know if I need to convert anything beforehand

Answer 4

It's given in seconds, divide it by 3600 to get in hours