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GreenLRan
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[SOLVED] Orbit Question
A satellite is in a circular Earth orbit of altitude 400km. Determine the new perigee and apogee altitudes if the satellite on-board engine gives the satellite a radial (outward) component of velocity of 240m/s.
Answers: Z_apogee = 621km, Z_perigee = 196km
mu = 398600km^3/sec^2
h (specific angular momentum) = r*v_perpendicular
a = semimajor axis of ellipse
2*a = perigee radius + apogee radius
v^2/2 + mu/r = -mu/(2*a) ( specific energy equation )
e= eccentricity of the orbit
theta = the true anomaly angle ( angle between eccentricity vector and the position vector)
r_perigee = a(1-e)
radius of the Earth = 6378km
r = h^2/mu * (1+e*cos(theta))
V_cir = sqrt(mu/r)
I started off by getting the magnitude of the velocity (v_perp^2 + v_radial^2)^.5
then calculated the escape velocity to see if it would be a closed orbit.
I found that it was still an ellipse (i also calculated the specific energy and found that it was negative... thus being an ellipse) from the specific energy i got 'a' ( -mu/(2*a) )
and using rp = a(1-e) = h^2/mu*(1+e)^-1 (at theta = 0 for rp)
however... when i did this I used the same h that i got from the satellite being a circular orbit.. which i am unsure if that was incorrect to do..
finally after solving for e, i plugged it back into rp = a(1-e) to get ~ 6543km... then the altitude zp = 6543- 6378 = 165km.. which is incorrect... can anyone help?? Thanks.
Homework Statement
A satellite is in a circular Earth orbit of altitude 400km. Determine the new perigee and apogee altitudes if the satellite on-board engine gives the satellite a radial (outward) component of velocity of 240m/s.
Answers: Z_apogee = 621km, Z_perigee = 196km
Homework Equations
mu = 398600km^3/sec^2
h (specific angular momentum) = r*v_perpendicular
a = semimajor axis of ellipse
2*a = perigee radius + apogee radius
v^2/2 + mu/r = -mu/(2*a) ( specific energy equation )
e= eccentricity of the orbit
theta = the true anomaly angle ( angle between eccentricity vector and the position vector)
r_perigee = a(1-e)
radius of the Earth = 6378km
r = h^2/mu * (1+e*cos(theta))
V_cir = sqrt(mu/r)
The Attempt at a Solution
I started off by getting the magnitude of the velocity (v_perp^2 + v_radial^2)^.5
then calculated the escape velocity to see if it would be a closed orbit.
I found that it was still an ellipse (i also calculated the specific energy and found that it was negative... thus being an ellipse) from the specific energy i got 'a' ( -mu/(2*a) )
and using rp = a(1-e) = h^2/mu*(1+e)^-1 (at theta = 0 for rp)
however... when i did this I used the same h that i got from the satellite being a circular orbit.. which i am unsure if that was incorrect to do..
finally after solving for e, i plugged it back into rp = a(1-e) to get ~ 6543km... then the altitude zp = 6543- 6378 = 165km.. which is incorrect... can anyone help?? Thanks.