Question

In: Physics

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee...

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee of its orbit it is 500 km above the earth surface at the high point or apogee it is 5000km above the earth surface
A) what is the period of the spacecraft orbit!?
B) using conservation of angular momentum find the ratio of the spacecraft speed at perigee to its speed apogee!?
C) using conservation of energy find the speed at perigee and the speed at apogee!?
D) it is necessary to have the spacecraft escape from the earth completely if the spacecraft rockets are fored at perigee by how much would be the speed have to be increased to achieve this!?
E) what if the rockets were fire at apogee?
F) which point is more efficient to use?

Solutions

Expert Solution

Given Data

hp = height of perigee = 500 km = 500000 m

ha = height of apogee = 5000 km = 5000000 m

We have

Re = radius of earth = 6380 km = 6380000 m

M = mass of the earth = 5.98*1024 kg

G = 6.67*10-11 Nm2/kg2

Solution :-

A) We know that semi-major axis is a = (1/2)[(Re + hp)+(Re +ha)]

                                                 = 9.13*106 m

We know that time period of revolution of the spacecraft is T= 2πa3/2/√GM

Then T = 2π*(9.13*106 )3/2/√6.67*10-11 *5.98*1024

            = 8.67*103 sec

B) We know that according to conservation of angularmomentum

    mvprp =mvara

vp = spacecraft's speed at perigee

va=spacecraft's speed at apogee

Then vp /va =ra/rp = (Re+ha)/(Re +hp)  

                          = 6380+5000 / 6380+500

                          = 1.654  

C) By conservation of energy

Now total energy at perigee is Tp =(1/2)mvp2 - GMm/rp

    total energy at apogee isTa = (1/2)mva2 -GMm/ra

then (1/2)mvp2 -GMm/rp   =  (1/2)mva2 -GMm/ra

      (1/2)va2- GM/ra =  (1/2)vp2 -GM/rp   

        (1/2)(rp/ra)2vp2-GM/ra                        =  (1/2)vp2 -GM/rp   

then vp2 = 2GM (ra/rp) / (rp+ra)

              = 2* 6.67*10-11 * 5.98*1024*1.654 /(11380*103+6880*103)

            = 7.22*107 (m/s)2

then vp = 8.5*103 m/s

we know that vp/va = 1.564

    then va = vp/1.564 = 5.433*103 m/s

genius_generous answered 2 years ago
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