Answer:
Approximately [tex]11121\; {\rm kg}[/tex] (rounded to the nearest whole number as required, assuming that planet Earth is a uniform sphere.)
Explanation:
Let [tex]G[/tex] denote the gravitational constant. [tex]G \approx 6.6743 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}[/tex].
Let [tex]M[/tex] denote the mass of planet Earth. Let [tex]m[/tex] denote the mass of the telescope. Let [tex]r[/tex] denote the distance between the telescope and the center of planet Earth.
Note the unit conversion for the distance [tex]r[/tex]:
[tex]\begin{aligned}r &= 6940 \; {\rm km} \times \frac{10^{3}\; {\rm m}}{1\; {\rm km}} \\ &= 6.940 \times 10^{6}\; {\rm m}\end{aligned}[/tex].
By Newton's Law of Universal Gravitation, the magnitude of the gravitational force between planet Earth and this telescope would be:
[tex]\begin{aligned}W &= \frac{G\, M\, m}{r^{2}}\end{aligned}[/tex].
Rearrange this equation to find the mass [tex]m[/tex] telescope in terms of [tex]G[/tex], [tex]M[/tex], and [tex]r[/tex]:
[tex]\begin{aligned}G\, M\, m &= W\, r^{2}\end{aligned}[/tex].
[tex]\begin{aligned}m &= \frac{W\, r^{2}}{G\, M}\end{aligned}[/tex].
Substitute in the value of [tex]G[/tex], [tex]M[/tex], and [tex]r[/tex]:
[tex]\begin{aligned}m &= \frac{W\, r^{2}}{G\, M} \\ &\approx \frac{9.21 \times 10^{4}\; {\rm N} \times (6.940 \times 10^{6}\; {\rm m})^{2}}{6.6743 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}} \times 5.98 \times 10^{24}\; {\rm kg}} \\ &\approx 11121\; {\rm kg}\end{aligned}[/tex].
