Question

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telescope is 9.21 × 104 N, and the mass of Earth is 5.98 × 1024 kg, what is the mass of the telescope? Round the answer to the nearest whole number

Answer 1

The answer is 11,121 kg

Answer 2

The gravitational attraction between the Earth and the Hubble telescope is given by:

$F=G\frac{M m}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the Hubble mass

r is the distance of the Hubble telescope from the earth's center

By rearranging the equation and substituting the numbers given by the problem, we find the mass of the telescope:

$m=\frac{F r^2}{GM}=\frac{(9.21 \cdot 10^4 N)(6.94 \cdot 10^6 m)^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(5.98 \cdot 10^{24}kg) } = 11121 kg$