Path followed by a space station.

[zorro]

Homework Statement

Consider a space station orbiting around the Earth in a circular orbit. If it fires its engine radially outward, will it follow a different circular path or an elliptical path?

The Attempt at a Solution

Since the thrust force is perpendicular to the motion, the angular momentum is conserved. So it should follow an elliptical path.

Is this reasoning correct?

 

[ideasrule]

Your answer is correct, but your reasoning is confusing. Why does the fact that angular momentum is conserved mean that it should follow an elliptical path?

 

[zorro]

V = √(GM/R)
L1 = m√(GMR)

V' = √(GM/R')
L2= m√(GMR')

where,
m - Mass of space station
V - Initial orbital velocity
V' - Final orbital velocity

L1 ≠ L2

So the path cannot be a circle. The other choice is only ellipse (as per the question)
I am interested in knowing the geniune reason though.

 

[zorro]

Can anybody else provide me a reason?

 

[hikaru1221]

Hint: The centripetal force always is: $$F = mv^2/\rho$$ where $$\rho$$ is the radius of curvature. If $$\rho = R$$ where $$R$$ is the distance between the station and the earth, then the orbit is circular around the earth.

You may want to prove this statement: for a fixed pair of planet M and mass m<<M orbitting around M, for each value of angular momentum L of m, there is one and only one circular orbit corresponding to L.

P.S.: If you want a genuine reason, look at the centripetal force. It's actually gravitational force. When does the gravitational force match *perfectly* with the centripetal force in the case of circular orbit?

 

[zorro]

You may want to prove this statement: for a fixed pair of planet M and mass m<<M orbitting around M, for each value of angular momentum L of m, there is one and only one circular orbit corresponding to L.

L=mvr=m√(GM/r)r=m√(GMr)
For every 'L' there is one and only one 'r'

When does the gravitational force match *perfectly* with the centripetal force in the case of circular orbit?

When they both are equal?