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pentagon [3]
3 years ago
8

The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS

T’s orbital period? The radius of Earth is 6.38 × 106 m.

Physics
2 answers:
deff fn [24]3 years ago
6 0

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

v=\frac{2\pi r}{T} (1)

where

2\pi r is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: v=7,750 m/s. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m

So, we can re-arrange equation (1) to find the orbital period:

T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s

Dividing by 60, we get that this time corresponds to 94 minutes.

laila [671]3 years ago
5 0

Answer:

5,630 s

Explanation:

I just got it right on ed2020

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