The tangential speed of the satellite above the Earth's surface is [tex]7.588 * 10^{3} m/s[/tex].
The tangential speed of a satellite at the given radius and time is calculated as follows:
[tex]v=\frac{2\pi *(150*10^{3}+6371*10^{3} }{90*60} \\v=7.588*10^{3} m/s[/tex]
Therefore, the tangential speed of the satellite above the Earth's surface is [tex]7.588 * 10^{3} m/s[/tex].
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The correct question is shown below:
Consider formula A to be v = and formula B to be v2 = G. Write the letter of the appropriate formula to use in each scenario. Determine the tangential speed of the moon given the mass of Earth and the distance from Earth to the moon. Determine the tangential speed of a satellite that takes 90 minutes to complete an orbit 150 km above Earth’s surface.