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What is the approximate tangential speed of an object orbiting Earth with a radius of 1.8 × 108 m and a period of 2.2 × 104 s?

7.7 × 10^–4 m/s
5.1 × 10^4 m/s
7.7 × 10^4 m/s
5.1 × 10^5 m/s

Respuesta :

MissPhiladelphia
If the period is 2.2 x 104 s, then the rotational speed is:
1 / 2.2x104 = 4.55x10-5 rad/s

We can solve for the tangential speed using the formula:
v = ωr
where ω is the rotational speed
v is the linear or tangential speed
r is the radius

So,
v = 4.55x10-5(1.8x108)
v = 5.1 x 104 m/s

Answer:

B

Explanation:

Edge 2020

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