Question

the orbital period of a satellite is 2×10⁶ s and its total radius is 2.5×10¹² m. the tangential speed of the satellite is m/s. (round to the nearest whole number, do not add any punctuation )

Explanation:

Step1: Calculate the circumference of the orbit

The formula for the circumference of a circle is \(C = 2\pi r\). Given \(r = 2.5\times10^{12}\text{ m}\), then \(C=2\times\pi\times(2.5\times 10^{12})\text{ m}\approx 1.5708\times 10^{13}\text{ m}\).

Step2: Calculate the tangential speed

The formula for tangential speed \(v=\frac{d}{t}\), where \(d\) is the distance (circumference of the orbit) and \(t\) is the time (orbital period). Given \(t = 2\times10^{6}\text{ s}\) and \(d\approx1.5708\times 10^{13}\text{ m}\), then \(v=\frac{1.5708\times 10^{13}}{2\times 10^{6}}\text{ m/s}\).
Using the rule of exponents \(\frac{a\times10^{m}}{b\times 10^{n}}=\frac{a}{b}\times10^{m - n}\), we have \(v=\frac{1.5708}{2}\times10^{13-6}\text{ m/s}=0.7854\times 10^{7}\text{ m/s}=7854000\text{ m/s}\approx7854000\text{ m/s}\) (rounded to the nearest whole number).

Answer:

7854000