Question

the hubble space telescope (hst) orbits 569 km above earth’s surface. given that earth’s mass is 5.97 × 1024 kg and its radius is 6.38 × 106 m, what is hst’s tangential speed? make sure to use correct significant figures.

Answer 1

Answer:

7,570 m/s

Explanation:

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Answer 2

Based on the calculations, the hst’s tangential speed is equal to 7570 m/s.

Given the following data:

Distance = 569 km.

Radius of Earth = $6.37\times 10^6\;m.$

Mass of Earth = $5.972\times 10^{24} kg$

Gravitational constant = $6.67 \times 10^{-11}$

How to calculate tangential speed.

Mathematically, tangential speed is given by this formula:

$V=\sqrt{\frac{GM}{r} }$

Where:

• r is the radius.

• G is the gravitational constant.

• M is the mass of Earth.

Substituting the given parameters into the formula, we have;

$V=\sqrt{\frac{6.67 \times 10^{-11} \times 5.972\times 10^{24}}{569 \times 10^3 + 6.37\times 10^6} }$

V = 7570 m/s.

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