Question

What is the magnitude of the gravitational force between the Earth and the Hubble Space Telescope (HST), which orbits the Earth? Assume the distance from the center of the Earth to the HST is 6,930 km, the mass of the Earth is 5.97 x 10²⁴ kg, the mass of the HST is 1.11 x 10⁴ kg, and G=6.67 x 10^-11 Nm²/kg².

Answer 1

As per newton's law of gravitation
F = GMm/(d^2)
Where G = 6.67 * 10^(-11) Nm^2kg^(-2) , d is the central distance in metres and M and m are masses of two bodies

now, F = (6.67*10^(-11) * 1.11*10^4 * 5.97*10^24)/(6.93 * 10^6 )^2
(Radius of earth + altitude of telescope = central distance)
=92035.77519 N

hence, the earth exerts a pull of 92035.77519 N on the telescope.

Answer 2

As per newton's law of gravitation
F = GMm/(d^2)
Where G = 6.67 * 10^(-11) Nm^2kg^(-2) , d is the central distance in metres and M and m are masses of two bodies

now, F = (6.67*10^(-11) * 1110 * 5.97*10^24)/(6930000)^2
(Radius of earth + altitude of telescope = central distance)
=9203.5 N