F=GMm/(R^2)
A=GM/(R^2)
Where M is the big Mass
G is "big G"
R is distance
Acceleration on a test particle in LEO due to the Earth = g times [6371 km/(LEO+6371 km)]^2
For LEO=400 km, A=0.885g
Acceleration on the same particle due to the Moon = g times (Mass Moon/Mass Earth) times (R Earth/R Moon)^2
For MM/ME = 1/81, and R Moon = 400,000 km,
A= 1/81 * (6371/400,000)^2
= 1/81 * 0.000254
= 3.13 *10^-6 g
Acceleration on the same particle due to the Sun = g times (Mass Sun/Mass Earth) times (R Earth/R Sun)
For MS/ME = 3.33*10^5, and R Sun = 139,600,000 km,
A= 6.03 *10^-4 g
Of course, the "test particle" can be an entire spacecraft.
This tells us that the Sun's gravity is ~200 times more important than the Moon's in determining the Earth's orbit.
Also, out of interest, the force of gravity on the "surface" of the Sun is ~ 7g
Space/Science » in reply to Re: You guys are both wrong:)
Newton's law...
The whole thread (34 posts)
- This looks like some bad news...
- Balony
- Implications of Hoffman's observation on the effect of the Moon on Earth
- Re: I havfe to say yes . . .
- You're wong...... Again.
- Re: You're wong...... Again. About the tides too.
- Re: You're wong...... Again. And technically so are you.
- Re: Nice to see you at least went to school . . .
- You're wong...... Again.
- Re: I havfe to say yes . . .
- Re: This looks like a load of horse manure
- Re: This looks like some bad news...
- I think it works both ways.
