You are designing a spacecraft intended to monitor a human expedition to mars (mass 6.42×1023kg, radius 3.39×106m). this spacecraft will orbit around the martian equator with an orbital period of 24.66 h, the same as the rotation period of mars, so that it will always be above the same point on the equator. what must be the radius of the orbit?

 h = 17 10⁶ m from surface of mars

Explanation:

For this exercise we will use Newton's second law where force is the force of universal gravitation

         F = m a

The acceleration is centripetal

         a = v² / r

         G m M / r² = m v² / r

The speed module is constant, so we use the uniform motion ratio

        v = d / t

Where the distance is the length of the circumference and the time is the period of the orbit

         d = 2π r

         v = 2π r / T

We replace

          G M / r² = (4π² r² / T) / r

           r³ = G M T² / 4π²

Let's reduce time to SI units

          T = 24.66 h (3600 s / 1 h) = 88776 s

Let's calculate

         r = ∛ 6.67 10⁻¹¹ 6.42 10²³ 88776² / 4π²

         r = ∛ 8.5485 10²¹ m

         r = 2,045 10⁶ m

This is the distance from the center of the planet, The height, which is the distance from the surface is

        r = + h

        h = r -

        h = 20.45 10 6 - 3.39 106

        h = 17 10⁶ m

he kicked it 10 feet straight up into the air? i think this works.

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