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# Determination of the mass of a planet from orbital dynamics

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A satellite of mass 830 kg orbits a planet of unknown mass at a distance of 29100 km from the planet's center. The orbital velocity of the satellite is 14200 m/s.

* What is the mass of the planet?
* How much would potential energy (PE) and kinetic energy (KE) of this satellite change as the satellite moved from a circular orbit of radius 29100 km to a circular orbit of radius 32010 km?

Consider G = 6.67 * 10^-11 N m^2 / kg^2 .

https://brainmass.com/physics/planets/determination-mass-planet-orbital-dynamics-395933

## SOLUTION This solution is FREE courtesy of BrainMass!

Satellite mass (m) in circular orbit with tangential velocity (v) of radius (r) from center of the planet has centripetal force

F = m*v^2/r (1)

To balance the satellite in orbit, this force equals the gravitational force given by (2).

F = GMm/r^2 (2)

Where M is the mass of the planet in question.

Equating (1) and (2) we get

m*v^2/r = G*M*m/r^2 (3)

Simplifying and cancelling a factor of m and r out both sides of (3) we get

v^2 = GM/r (4)

Mass of planet is on re-arrangement of (4)

M = r*v^2/G (5)

We are given r = 29,100 km = 2.91 x 10^7 m

v = 14,200 m/s = 1.42 x 10^4 m/s

G = 6.67 x 10^-11 m^2/kg^2

Putting in figures to (5)

M = 2.91 x 10^7 x (1.42 x 10^4)^2/6.67 x 10^-11 = 8.8 x 10^25 kg

If the satellite moved to a new orbit of radius R = 32,010 km = 3.21 x 10^7 m

New velocity (V^2) squared is given by (4) as

V^2 = GM/R = 6.67 x 10^-11 x 8.8 x 10^25/3.21 x 10^7 m2/s2

V^2 = 1.83 x 10^8 m2/s2

Original Kinetic energy (KE) is given by

KE(at r) = 1/2 m*v^2 = 0.5 x 830 x 14200 = 83.68 GJ

New Kinetic energy is given by

KE(at R) = 1/2 m*V^2 = 0.5 x 830 x 1.83 x 10^8 = 75.95 GJ

Thus going from r to R (29,100 km orbit to 32,010 km circular orbit) the satellite loses

83.68 - 75.95 = 7.73 GJ of KE

The way the Potential Energy (PE) changes is the opposite of this so the satellite gains 7.73 GJ of PE.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!