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The gravitational force between two bodies of mass m1 and m2 should be F = G*m1*m2 / r².

The mass of the moon = 7.346 *10²² kg, G = 6.67 *10^-11 m³/kgs²,and you stated that r = 100,000 km (the radius of the moon is only ~2,000 km so we won't worry that we are measuring r from the surface not the center.)

From this, we can say the force on a 1 kg object would be 0.00049 N, or 0.000011 lbs. In other words, this would be 20,000 x less than the gravity we feel on the surface of the earth.

This should make sense, the mass of the moon is ~80 times smaller than that of the earth. The radius of the earth is ~6,500 km, so your r is 16 times larger.

Thus from before for F=Gm1*m2/r², if we decrease m2 by a factor of 80, and increase r by a factor of 16, the new force should be (1/80) / (16)² = 1/20,000 what we feel on earth.

The acceleration due to gravity at distance R from the center of a spherical body is GM/R² where G is the universal gravitational constant, M is mass of moon or planet, and R is the distance from the center of the body in a radial direction.

At the surface of the body it would be the body's radius, so for Earth it is g=G*Mearth/Rearth², but if one was 1000 miles above the Earth surface, it would be Rearth + 1000 miles , etc.

Let’s take a look at Newton’s law of universal gravitation to help answer your question.

Newton’s law states that the force (F) enacted on two bodies interacting with each other due to gravity is the product of the gravitational constant (G), the masses of the two bodies (m1 and m2) and the inverse of the distance squared between their center of mass (1/r²). The complete equation is: F = G*m1*m2/r².

If we take out the mass of the variable body (i.e. keep the mass of the moon in the equation), the equation reduces to the acceleration (a) due to the gravitational force of the remaining body. You can verify that the acceleration at the surface of the Earth should be 9.8 m/s² (one G) by using these numbers (from hypertextbook.com; make sure units correctly cancel out!):

G = 6.7X10^-11 m³/kg/s

m_earth = 6X10²⁴ kg

r = (diameter_earth)/2 = (12800 km) / 2 = 6400 km = 6.4X10⁶ m

Thus, a = G*m_earth/r² = 9.81 m/s² (correct!)

So, now we can calculate the acceleration of an object 100,000 km from the surface of the moon. We need to make sure to add the distance from the center of the moon to this 100,000 km (numbers again from hypertextbook.com):

m_moon = 7.5X10²² kg

r = 100,000 km + (diameter_moon)/2 = 100,000 km + (3480 km)/2 = 1.02X10⁸ m

Thus, a = G*m_moon/r² = 4.8X10^-4 m/s²

The results should be expected; we are looking at the gravitational effects of something that is very, very far away from the moon (about 1/4 the distance from the Earth to the moon). But while the numbers seem very small, they are not negligible. The small shifts in gravitational force are indeed enough to cause changes on Earth such as the rise and fall of tides due to the moon (and Sun).

The inverse square law is a very interesting one in that it shows up everywhere in physics. As applied to gravity the law can be written as such: F = G{m1*m2}/r².

F is the force in newtons, G is the gravitational constant (6.674×10^-11 N m² kg^-2)

m1 and m2 are the masses of the two things being attracted, in your case, the moon with a mass of 7.347 x 10²² kg, and some other mass (in kg) and the distance between the masses, in meters.

The reason you need two masses is because gravity can only be felt by its effect on matter (like all fundamental forces) and since gravity only works on mass, we can only measure it based on its effect on something with mass. Two massive objects will attract each other with equal force. So, that is to say, when you are falling the earth's gravity is exerting a certain force on you, but you are exerting the exact same force on the earth. The difference is that the earth is so large that it really doesn't move very much at all. (This is because F = MA, force equals mass times acceleration. If the force is the same then your acceleration will be much higher than the earth's because your mass is so much lower than the earth's mass).

So the problem you may be having could be the value you use for G, could be the second mass that you use, or it might be the units you're using for all the values. I hope that helps.