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

The centripetal force that must act on the moon is approximately 2.729 × 10^20 N. This force is greater than the gravitational force that the earth exerts on the moon at the same distance.

Step-by-step explanation:

To calculate the centripetal force that must act on the moon, we can use the formula F = (mv^2)/r, where m is the mass of the moon, v is the velocity of the moon, and r is the radius of the orbit. The velocity of the moon can be calculated using the formula v = 2πr/T, where T is the time required for one revolution. Plugging in the given values, we find that the centripetal force is approximately 2.729 × 10^20 N.

The gravitational force that the earth exerts on the moon at the same distance can be calculated using the formula F = (GmM)/r^2, where G is the gravitational constant, M is the mass of the earth, m is the mass of the moon, and r is the distance between them. Plugging in the given values, we find that the gravitational force is approximately 1.982 × 10^20 N. Therefore, the centripetal force is greater than the gravitational force.

Answer 2

Step-by-step explanation:

It is given that,

Mass of moon,
`m=7.35* 10^(22)\ kg`

Radius of circle,
`r=3.82* 10^(5)\ km=3.82* 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=(2\pi r)/(T)`

`v=(2\pi * 3.82* 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=(mv^2)/(r)`

`F=(7.35* 10^(22)\ kg* (1017.57\ m)^2)/(3.82* 10^(8)\ m)`

`F=1.99* 10^(20)\ m/s^2`

So, the centripetal force that must act on the moon is
`1.99* 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* 10^(20)\ m/s^2`. Hence, this is the required solution.