Question

Calculate the magnitude of the gravity acceleration on the surface of the moon and again at a point 1000 km above the surface of the moon; ignore the gravity effects of the earth. The moon has a mean radius of 1740 km and a mass of 7.4x10^22 kg.

Answer 1

Newton's Universal Law of Gravitation states that between 2 masses, m and M, if the distance is R, then the following equation holds true\[Force=G \frac{ Mm }{ R^2 }\] where G is a gravitational constant.
By Newton's Second Law of Motion, \[Force=ma\] where m is a mass and a is the acceleration.

Answer 2

Set them equal to get\[a=G \frac{ M }{ R^2 }\]

Answer 3

For this problem, M is the mass of the moon.

Answer 4

For the first case, R would be the distance from surface to the center of the moon, or the mean radius of the moon.

Answer 5

So the acceleration there would be\[a=G \frac{ 7.4*10^{22} }{ 1740^2 }\]
G=6.67384*10^-11. You can round it if you want

Answer 6

For the second case, R would be the distance from where the object is to the center of the moon.

Answer 7

So R would be 2740 in this case.
Substitute the values in to get the second answer.

Answer 8

Any questions?

Answer 9

wait

Answer 10

Is G = gravitational constant of moon?

Answer 11

and what is the unit of that G?

Answer 12

G is called the Universal Gravitational Constant.
It's a constant that never changes throughout the universe.

unit of G is
\[N*(m/kg)^2\]or
\[m^3kg^{-1}s^{-2}\]

Answer 13

any other questions?

Answer 14

Ohhh. Okay so it must be 1,740,000 m. :))
Okay. I hope you're always online. I have lots of questions. :)) "Practice questions*

Answer 15

Goodbye!

Answer 16

Very good I learned new thing from you! :))