# Moon gravity

Ockonal

## Homework Statement

Mass of the Moon is nearly less in 100 times than Earth's, and the diametr is less in 4 times.
Find the acceleration of free falling at the Moon.

F = G×((M×m)/R²)
g = 9.8N

## The Attempt at a Solution

M_moon = M_earth / 100
D_moon = D_earth / 4
R_moon = D_moon / 2 = D_earth / 8

g_earth = G×(M_earth/R_earth²)
g_moon = G×(M_moon / R_moon²)
=>
M_earth = (g×R_earth²) / G
M_moon = (g×R_earth²) / G / 100
...
g_moon = g_earth / 25
But the answer is wrong. I think my mistake is in diameter passing and converting it into radius. Help me, please/

Staff Emeritus
Your mistake is that you dropped something. You were fine up to here:
g_earth = G×(M_earth/R_earth²)
g_moon = G×(M_moon / R_moon²)
Given that, what is g_moon / g_earth?

Ockonal
so:
g_moon / g_earth =

M_moon × R_earth²
-------------------
M_earth × R_moon²

Now I'm passing data instead of *_moon:

(M_earth/100) × R_earth²
----------------------
M_earth × (D_earth / 4)

How to make it easier?

Ockonal
so:
Code:
g_moon / g_earth =

M_moon × R_earth²
-------------------
M_earth × R_moon²
Now I'm passing data instead of *_moon:
Code:
(M_earth/100) × R_earth²
----------------------
M_earth × (D_earth / 4)
How to make it easier?

-----
I did this:

Code:
  M_earth × R_earth²
----------------------
100
----------------------
M_earth × 2 × R_earth
----------------------
8
=>
R_earth / 25

But R_earth shouldn't be presented.

Staff Emeritus
If the diameter of the Moon is roughly 1/4 of the diameter of the Earth, what can you say about the radius of the Moon in terms of the radius of the Earth?

Ockonal
If the diameter of the Moon is roughly 1/4 of the diameter of the Earth, what can you say about the radius of the Moon in terms of the radius of the Earth?

D_moon = 1/4 D_Earth
R_moon = (1/4 D_Earth)/2

I'm confused :(

Staff Emeritus
You just expressed the Moon's radius in terms of the Earth's diameter. I asked you to express the Moon's radius in terms of the Earth's radius. So, try again, please.

Once you have that figured out, go back to computing g_moon / g_earth.

cupid.callin
you know that g=Gm/r2

now for earth

g = GM/R2

for moon, g' = GM'/R'2

you know that M' = M/100 and D' = D/4

so R' = ?

Ockonal
You just expressed the Moon's radius in terms of the Earth's diameter. I asked you to express the Moon's radius in terms of the Earth's radius. So, try again, please.

Once you have that figured out, go back to computing g_moon / g_earth.

As I understand:
D/4 = Moon's diametr, so:
R/4 + R/4 = D/4
So radius is smaller in 4 times?

cupid.callin
yes!!!

now just substitute the values

Ockonal
now just substitute the values
Didn't understand right, maybe:
Code:
g_moon = G × ( M_earth / 25 × R_earth² )
g_earth = G × ( M_earth / R_earth² )

g_moon / g_earth =
M_eath × R_earth²
-----------------------   =
M_earth × 25 × R_earth²

1/25

cupid.callin
the ' ones are for moon

g = GM/R2

for moon, g' = GM'/R'2

you know that M' = M/100 and D' = D/4

so R' = R/4

Code:
g' = G(M/100)
_______
R/4

so ... g' = <something> GM/R[sup]2[/sup]
g' = <something> g

Ockonal
the ' ones are for moon
Code:
g' = G(M/100)
_______
R/4

so ... g' = <something> GM/R[sup]2[/sup]
g' = <something> g

Sorry, I can't understand from where did you get GM/R² in
Code:
g' = G(M/100)
_______
R/4

cupid.callin
sorry i cant give answer ,,, got 2 warnings for it

i just rearranged the eqn and dividing the numbers i can

separate GM/R² ... and as you know from eqn of earth GM/R² = g

eqn of earth ... sounds fascinating!!