What would be the orbital speed and period of a satellite in orbit 0.82 ✕ 108 m above the earth?

Newton's second law and the law of universal gravitation of attraction allows to find the result for the speed of the satellite in Earth orbit is:

            v = 2.21 10³ m / s

Newton's second law gives a relationship between force, mass and acceleration of bodies.

          F = ma

Where F is the force, m the mass and the acceleration.

The law of universal gravitation says that the force between two bodies is proportional to the mass and inversely proportional to the distance between them.

Where G is the universal gravitational constant, M and m the masses of the bodies and r the distance between them.

We substitute

as the orbit is circular, the acceleration is cenripetal, see attached.

Where v is the speed of the satellite.

Let's substitute.

The distance from the inside of the planet to the satellite is r = 0.82 10⁸ m, the mass of the planet is tabulated M + 5.98 10241 kg. Let's calculate.

           v = 2.21 10³ m / s

In conclusion, using Newton's second law and tlaw of universal gravitation of attraction we can find the result for the speed of the satellite in Earth orbit is:

            v = 2.21 10³ m / s

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total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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answer: a. same unit

explanation: