Question

what is the approximate tangential speed of an object orbiting earth with a radius of 1.8×10^8 m and a period of 2.2×10^4 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

Explanation:

1. First, recall the formula for the circumference of a circle (the distance traveled in one - orbit):

• The formula for the circumference of a circle is \(C = 2\pi r\), where \(r\) is the radius of the orbit. Given \(r = 1.8\times10^{8}\text{ m}\), then \(C = 2\pi(1.8\times10^{8})\text{ m}\).
• The tangential speed \(v\) of an object in circular - motion is given by the formula \(v=\frac{d}{t}\), where \(d\) is the distance traveled and \(t\) is the time taken. In the case of circular motion, the distance traveled in one period \(T\) is the circumference of the circle, so \(v=\frac{2\pi r}{T}\).

1. Then, substitute the given values:

• We know that \(r = 1.8\times10^{8}\text{ m}\) and \(T = 2.2\times10^{4}\text{ s}\).
• \(v=\frac{2\pi\times(1.8\times10^{8})}{2.2\times10^{4}}\text{ m/s}\).
• First, calculate \(2\pi\times1.8\times10^{8}\approx2\times3.14\times1.8\times10^{8}= 11.304\times10^{8}\text{ m}\).
• Then, \(\frac{11.304\times10^{8}}{2.2\times10^{4}}=\frac{11.304}{2.2}\times10^{8 - 4}\text{ m/s}\).
• \(\frac{11.304}{2.2}\approx5.14\), so \(v\approx5.14\times10^{4}\text{ m/s}\approx5.1\times10^{4}\text{ m/s}\).

Answer:

5.1×10⁴ m/s