Circular Orbit of A Satellite

[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.

 

[rwisz]

= -GMe*m/2r

Etot = -GMe*m/2r

In response to the content, as a scientist, I would say that you have done a good job in understanding the basic concepts and equations related to circular orbit of a satellite. Your calculations and equations seem to be correct. To answer your question about the tangential speed, it depends on the reference point you choose. If you choose the center of the Earth as your reference point, then you should use the radius of the Earth in the R part of the equation for tangential speed. If you choose the satellite's position as your reference point, then you should use the distance from the satellite to the center of the Earth. Both equations are correct, it just depends on the reference point you choose. Good job on your first question and keep up the good work in learning more about circular orbits and satellite motion.