Question

The moon's mass is 7.35 × 1022 kg, and it moves around the earth approximately in a circle or radius 3.82 × 105 km. The time required for one revolution is 27.3 days. Calculate the centripetal force that must act on the moon. How does this compare to the gravitational force that the earth exerts on the moon at that same distance?

Answer 1

Explanation:

It is given that,

Mass of moon, $m=7.35\times 10^{22}\ kg$

Radius of circle, $r=3.82\times 10^{5}\ km=3.82\times 10^{8}\ m$

The time required for one revolution is 27.3 days, t = 27.3 days

1 day = 86400 seconds

27.3 days = 2358720 seconds

Let v is the speed of moon around the circular path. It is given by :

$v=\dfrac{2\pi r}{T}$

$v=\dfrac{2\pi \times 3.82\times 10^{8}\ m}{2358720\ s}$

v = 1017.57 m/s

Let F is the centripetal force acting on the moon. It is given by :

$F=\dfrac{mv^2}{r}$

$F=\dfrac{7.35\times 10^{22}\ kg\times (1017.57\ m)^2}{3.82\times 10^{8}\ m}$

$F=1.99\times 10^{20}\ m/s^2$

So, the centripetal force that must act on the moon is $1.99\times 10^{20}\ m/s^2$. The gravitational force that the earth exerts on the moon at that same distance is also equal to $1.99\times 10^{20}\ m/s^2$. Hence, this is the required solution.