Question

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_avg_earth = 6.38e6m

Answer 1

7020.117 m/s

Step-by-step explanation:

Answer 2

Answer: 7020.117 m/s

Step-by-step explanation:

The velocity of a satellite describing a circular orbit is constant and defined by the following expression:

`V=\sqrt{G(M)/(R)}` (1)

Where:

`G=6.674({10}^(-11))\frac{N{m}^(2)}{{kg}^(2)}` is the gravity constant

`M_(Earth)=5.97{10}^(24)kg` the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

`R=r_(Earth)+h=8080000m` the radius of the orbit (measured from the center of the planet to the satellite).

This means the radius of the orbit is equal to the sum of the average radius of the Earth
`r_(Earth)` and the altitude of the satellite above the Earth's surface
`h`.

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

`V=\sqrt{(6.674({10}^(-11))\frac{N{m}^(2)}{{kg}^(2)})\frac{5.97{10}^(24)kg}{8080000m}}`

`V=7020.117(m)/(s)`