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. Kg.

Answers

Answer 1

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].

Answer 2

Answer:

11,121 kg

Explanation:

The Center Of The Hubble Space Telescope Is 6940 Km From Earths Center. If The Gravitational Force Between

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