Question

Suppose a 1300-kg Tesla ends up orbiting some faraway planet of
mass 2.8 * 10^{24} kg. If that planet's radius is 6,200 km,
and if the Tesla arrived with a velocity relative to the planet of
5,200 m/s, and if the orbit were circular, how high above the
planet's ground, in kilometers, would the Tesla orbit?

use G = 6.674*10^{-11} Nm^{2}/kg^{2}

Answer #1

Given that, mass of planet, **M = 2.8 x 10 ^{24}
kg**

Radius of planet, **R = 6200 km**

Orbital speed, **v = 5200 m/s**

Let **"h"** be the height from planet's ground.

We know that, orbital speed is given by:

**v = sqrt
(GM/(R+h))**

**Therefore, height is given by:**

**R + h = (GM) /
(v ^{2})**

**h = [(GM) / (v ^{2})] -
R**

**h = [(6.674 x 10 ^{-11}
x 2.8 x 10^{24} ) / (5200^{2})] - 6200
km**

**h = (6910.95 x 1000 m) - 6200
km**

**h = (6910.95 km) - 6200
km**

**h = 710.95 km**
---------- (**Answer**)

_{===========================================================}

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