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. Kg.

Answer 1

11,121 kg

Step-by-step explanation:

Answer 2

Approximately
`11121\; {\rm kg}` (rounded to the nearest whole number as required, assuming that planet Earth is a uniform sphere.)

Step-by-step explanation:

Let
`G` denote the gravitational constant.
`G \approx 6.6743 * 10^(-11)\; {\rm N \cdot m^(2) \cdot kg^(-2)}`.

Let
`M` denote the mass of planet Earth. Let
`m` denote the mass of the telescope. Let
`r` denote the distance between the telescope and the center of planet Earth.

Note the unit conversion for the distance
`r`:

`\begin{aligned}r &= 6940 \; {\rm km} * \frac{10^(3)\; {\rm m}}{1\; {\rm km}} \\ &= 6.940 * 10^(6)\; {\rm m}\end{aligned}`.

By Newton's Law of Universal Gravitation, the magnitude of the gravitational force between planet Earth and this telescope would be:

`\begin{aligned}W &= (G\, M\, m)/(r^(2))\end{aligned}`.

Rearrange this equation to find the mass
`m` telescope in terms of
`G`,
`M`, and
`r`:

`\begin{aligned}G\, M\, m &= W\, r^(2)\end{aligned}`.

`\begin{aligned}m &= (W\, r^(2))/(G\, M)\end{aligned}`.

Substitute in the value of
`G`,
`M`, and
`r`:

`\begin{aligned}m &= (W\, r^(2))/(G\, M) \\ &\approx \frac{9.21 * 10^(4)\; {\rm N} * (6.940 * 10^(6)\; {\rm m})^(2)}{6.6743 * 10^(-11)\; {\rm N \cdot m^(2) \cdot kg^(-2)} * 5.98 * 10^(24)\; {\rm kg}} \\ &\approx 11121\; {\rm kg}\end{aligned}`.