Question

Suppose a 1300-kg Tesla ends up orbiting some faraway planet of mass 2.8 * 10²⁴ 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 Nm²/kg²

Answer 1

Given that, mass of planet, M = 2.8 x 10²⁴ 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) / (v²)

h = [(GM) / (v²)] - R

h = [(6.674 x 10^-11 x 2.8 x 10²⁴ ) / (5200²)] - 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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