What Speeds Are Required for a Satellite's Circular Orbit?

[MaroonR]

This is my first question here, so I'm a little new at this. I've really learned a lot from this forum in the past. Here goes:


1. Suppose you are responsible for a no-engine rocket project which carries a new satellite into space

a. What is the the minimum initial speed of the rocket to reach the height h from the Earth surface? Use Re for radius of earth, the Earth's mass Me, and the univsersal gravitational constant G. Neglect Air resistance)

b. After reaching height h, what is the tangential speed needed to make a circular orbit at height h

c. What is the total energy of the satellite?





Equations: K = .5mv^2, U = -G(Me*m/Re, Fc = m*v^2/r, K_circularorbit = GMem/2r


a: .5mv^2 - GMem/Re = - GMem/Re + h (solve for v)
solution: v = sqrt(2GMe((1/Re) - (1/(Re + h)

b: I have a question here. When using the centripetal acceleration for the earth, should you include the radius of the Earth in R part of v^2/r? If so, the equation for the tangential speed would simply be:

Fc = Fg

((mv^2)/(r+h)) = GMe*m/((r+h)^2)

v = sqrt(GM*m/(r+h), where Fc = mv^2/(r + h)

Otherwise, it would be: v = sqrt(GM*m/(r((r+h)^2)), where Fc = mv^2/r

which one of those is the correct usage?

c.

Etot = K + U

GMe*m/2r - GMe*m/r

 

[gneill]

Hi MaroonR. Welcome to Physics Forums.

Your solution for part a. looks okay.

In part b., I think you'll find that the small m's cancel (the mass of the satellite), and you should be left with v = sqrt(G*Me/(Re + h)). You can use the acceleration due to gravity and the centripetal acceleration rather than the force, and the mass of the satellite won't even appear:

A_c = A_g ==> v²/(R_e + h) = G*M_e/(R_e + h)²

Thus v = sqrt(G*M_e/(R_e + h))

For part c., you're on the right path. Let r = Re + h and simplify.

 

[MaroonR]

So my equation for centripetal acceleration SHOULD include the radius of the Earth in addition to the height of the satellite?

 

[gneill]

So my equation for centripetal acceleration SHOULD include the radius of the Earth in addition to the height of the satellite?

Certainly. It's orbiting the center of the planet, not some point on its surface.