- #1
Like Tony Stark
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- Homework Statement
- A spacecraft designed to reach Mars has mass=2000 kg. The first stage moves in an almost parabolic orbit. When it reaches the closest point (5 times Mars radius), it starts moving in a circular orbit. After that, it moves in an elliptical orbit and lands. Determine
A) what magnitudes of the spacecraft are conserved
B) angular momentum in each orbit
C) velocity when it reaches the closest point in the parabolic orbit
D) energy needed to change to the circular orbit, and then to the elliptical one.
E) velocity when it lands
- Relevant Equations
- ##E_M=E_C+E_P##
##L=rxv##
Tell me if I'm right:
A) Angular momentum is conserved because there are no external torques. Linear momentum isn't conserved because gravity is acting on the spacecraft . Mechanical energy isn't conserved because it has to change between different orbits.
B) Parabolic orbit: ##L=mv_1.5r_{Mars}##
Circular: ##L=mv_2 5r_{Mars}##
Elliptical: ##L=mv_3r_{Mars}##
I don't know if I should consider ##v_1## as given because that's what I'm asked in C. If I should't then I don't know how to calculate ##L##.
D) I know that ##E_M=E_C+E_P##, so I have to do ##\Delta E##. Kinetic energy is simple, but what about potenial energy? From circular to elliptical we have ##E_{cir}=-GmM/5r_m## and ##E_{elip}=-GmM/2(r_a+r_p)##, where ##r_p=r_{Mars}## but what about ##r_a##?
E) That's just the velocity at perigee.
A) Angular momentum is conserved because there are no external torques. Linear momentum isn't conserved because gravity is acting on the spacecraft . Mechanical energy isn't conserved because it has to change between different orbits.
B) Parabolic orbit: ##L=mv_1.5r_{Mars}##
Circular: ##L=mv_2 5r_{Mars}##
Elliptical: ##L=mv_3r_{Mars}##
I don't know if I should consider ##v_1## as given because that's what I'm asked in C. If I should't then I don't know how to calculate ##L##.
D) I know that ##E_M=E_C+E_P##, so I have to do ##\Delta E##. Kinetic energy is simple, but what about potenial energy? From circular to elliptical we have ##E_{cir}=-GmM/5r_m## and ##E_{elip}=-GmM/2(r_a+r_p)##, where ##r_p=r_{Mars}## but what about ##r_a##?
E) That's just the velocity at perigee.