The Hubble Space Telescope orbits the Earth at an approximate altitude of 612 km. Its mass is 11,100 kg and the mass of the Earth is 5.97×1024 kg. The Earth's average radius is 6.38×106 m. What is the magnitude of the gravitational force that the Earth exerts on the Hubble?
To calculate the magnitude of the gravitational force that the Earth exerts on the Hubble Space Telescope, we can use Newton's Law of Universal Gravitation. The formula is as follows:
F = (G * m1 * m2) / r^2
Where:
F is the gravitational force
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
m1 is the mass of one object (mass of the Earth)
m2 is the mass of the other object (mass of the Hubble)
r is the distance between the centers of the two objects (radius of the Earth plus the altitude of the Hubble)
In this case, the mass of the Earth (m1) is given as 5.97×10^24 kg, the mass of the Hubble (m2) is given as 11,100 kg, and the distance (r) is the sum of the radius of the Earth and the altitude of the Hubble, which is 612 km or 612,000 meters.
Now let's plug in these values into the equation:
F = (G * m1 * m2) / r^2
F = (6.67430 × 10^-11 * 5.97×10^24 * 11,100) / (6.38×10^6 + 612,000)^2
Calculating this expression will give us the magnitude of the gravitational force that the Earth exerts on the Hubble Space Telescope.
Your mass and radius of the Earth are not correct.
The Earth's radius is 6380 km .
F = Msat*Mearth*G/Rsat^2
= Msat* g * (Rearth/R)^2
Rsat = Rearth + 612 km = 6992 km
Re/R = 0.9125
g = 9.81 m/s^2
Hubble's weight in that orbit is
(0.9125)^2 = 83.3% of its weight on the Earth's surface.