An astronaut stands at a position on the Moon such that Earth is directly overhead and releases a Moon rock that was in her hand.

(c) What is the gravitational force exerted by the Earth on the same 2.7-kg rock resting on the surface of the Moon?
answer in mN

I keep calculating this but am getting it wrong please help
((6.67x10^-11Nm^2) / kg^2 x 2.7kg x (7.35x10^22kg) / (3.8226x10^8m)^2).

gives me about 9.06x10^-5 N so mN its about .09 mN but its wrong.

F =G•m•M(E)/[R(EM) –R(M)]²,

where
the gravitational constant G =6.67•10^-11 N•m²/kg²,
m =2.7 kg,
Earth’s mass is M(E) = 5.97•10^24 kg,
Distance between the centres of the Earth
and the Moon
R(EM) =384467000 m.
Moon’ radius R(M) =1737000 m.
My result is F = 7.34•10^-3 N

To calculate the gravitational force exerted by the Earth on the Moon rock, you can use the formula:

F = (G * m1 * m2) / r^2

Where:
F is the gravitational force
G is the gravitational constant (approximately 6.67 x 10^-11 Nm^2/kg^2)
m1 is the mass of the Moon rock (2.7 kg in this case)
m2 is the mass of the Earth (approximately 5.97 x 10^24 kg)
r is the distance between the center of the Earth and the center of the Moon (approximately 3.8226 x 10^8 m)

Let's plug in the values:

F = (6.67 x 10^-11 Nm^2/kg^2 * 2.7 kg * 5.97 x 10^24 kg) / (3.8226 x 10^8 m)^2

Now, let's calculate it step by step:

1. Calculate the denominator:
(3.8226 x 10^8 m)^2 = 1.4601 x 10^17 m^2

2. Multiply the masses:
2.7 kg * 5.97 x 10^24 kg = 1.6149 x 10^25 kg

3. Evaluate the numerator:
6.67 x 10^-11 Nm^2/kg^2 * 1.6149 x 10^25 kg = 1.0769233 x 10^15 Nm^2

4. Calculate the final gravitational force:
F = (1.0769233 x 10^15 Nm^2) / (1.4601 x 10^17 m^2)

F ≈ 7.3837 x 10^-3 N

To convert this to millinewtons (mN), you multiply the result by 1000.

7.3837 x 10^-3 N * 1000 = 7.3837 mN

Therefore, the gravitational force exerted by the Earth on the 2.7 kg Moon rock resting on the surface of the Moon is approximately 7.3837 mN.