Why Is the Mechanical Energy of a Satellite Negative?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
songoku
Messages
2,521
Reaction score
395

Homework Statement


the gravitational force exerted on a body of mass m by the Earth is GmM / r2

1. Express the speed of an artificial satellite which carries out uniform circular motion at height R from the surface of the Earth in terms of g and R

2. express the mechanical energy of the artificial satellite of (1) in terms of g, m, and R, where m is the mass of the artificial satellite and the potential energy is assumed to be zero when the distance r is infinite.

Homework Equations


Em = Ep + Ek


The Attempt at a Solution


1. done [ans : v =√(gR/2) ]

2.
Em = Ep + Ek = 0 + 1/2 mv2 = 1/2 m (gR/2) = mgR / 4

but the answer is - mgR / 4. Why is there negative sign?


thanks
 
Physics news on Phys.org
It says, "assuming the potential energy is 0 when R is infinite". You assumed the potential energy was zero at the distance R. So there is a potential energy term you forgot to add in. As to why it is negative, the satellite has more potential energy when it is further away. This is clear if you think about an object at rest at infinity and falling inward. As it gains kinetic energy, it needs to lose potential energy so that total energy is conserved. So if the total energy is zero at infinity (since PE=0, and KE=0 since v=0), as it accelerates inward and KE increases positively, PE must increase negatively.
 
phyzguy said:
It says, "assuming the potential energy is 0 when R is infinite". You assumed the potential energy was zero at the distance R. So there is a potential energy term you forgot to add in. As to why it is negative, the satellite has more potential energy when it is further away. This is clear if you think about an object at rest at infinity and falling inward. As it gains kinetic energy, it needs to lose potential energy so that total energy is conserved. So if the total energy is zero at infinity (since PE=0, and KE=0 since v=0), as it accelerates inward and KE increases positively, PE must increase negatively.

Oh, you're right. Got it now. Thanks a lot :smile: