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?

Answer 1

V (HST’s tangential speed) =  7570 m/s

Answer 2

The tangential speed of the Hubble Space Telescope (HST) is 7,569.7 m/s.

The given parameters;

• distance of the orbit, r = 569 km
• mass of the Earth, m = 5.97 x 10²⁴ kg
• radius of the Earth, r₂ = 6.38 x 10⁶ m

The gravitational force of attraction between the Earth and the HST is equal to the centripetal force of the HST.

$\frac{Gm_1M_e}{R^2 } = \frac{m_1 v^2}{R} \\\\\frac{GM_e}{R} = v^2\\\\v= \sqrt{\frac{GM_e}{R}} \\\\v = \sqrt{\frac{(6.67 \times 10^{-11}) \times (5.97\times 10^{24})}{(569,000 \ + \ 6.38\times 10^6)}} \\\\v = 7,569.7 \ m/s$

Thus, the tangential speed of the Hubble Space Telescope (HST) is 7,569.7 m/s.

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