Re: Surface Gravity to tell actual force one object has on another?

First of all, in my opinion, if your teacher can't tell you why your answer is wrong, the problem shouldn't have been assigned in the first place.....

But, to answer your question....

You are also confusing surface gravity with tidal gravity. In the former, only the planet's (or Moon's) mass and radius are involved. In the latter, both planet and moon have to be accounted for (M(planet) X M(moon)), as well as the distance between them.

Remember Newton's Law:

              agravity = G * M / r^2

for 1 body (surface gravity), or

          Fgravitational = [G * Mplanet * Mmoon]/d^2 

In this case, however, you don't even need to worry about units at all! Newton's Law says that F is proportional to 1/d^2. So if the Moon's distance increases by 2, it's force on Earth decreases by 1/2^2 = 1/4. It has 1/4 the effect on Earth at its new distance.

Additionally, tidal forces are a measure of differences in gravity, so tidal forces decrease as 1/d^3. Tides arise because the water on the near side of the Earth feels a stronger force than the water on the far side of the Earth.

So, if the Moon were twice its current distance, tidal forces would decrease by 1/2^3 = 1/8. Tidal forces would be 1/8 the current value at the new distance. The farther the Moon gets from Earth, the less its gravity affects the tides.

1. Mars Mission Launch Sequence

2. "Moons and Planets", W. K. Hartmann (1999), pages 57-59.