Question

prooblem 3!

Answer 1

3.

In Circular orbit transfer the lowest possible energy is spent at Hohmann Transfer orbit.

7500km Hohinann Transfer Orbit 4000km Planet

If standard gravitational parameter of the larger body is $GM = \mu = 6.67*10^{-11}*M_{earth}*1.148 = 4.57*10^{14}kg$ . (G is the universal gravitational constant, M is the mass of the large planet.)

And r1 is the initial orbit = $(r_{earth} *1.148) + 4000*10^3 = 1.13*10^7m$

r2 is the final orbit = $(r_{earth} *1.148) + 7500*10^3 = 1.48*10^7m$ , then,

$\Delta v_1$ required to transfer to the Hohmann transfer Orbit:

$\Delta v_1 ={\sqrt {\frac {\mu }{r_{1}}}}\left({\sqrt {\frac {2r_{2}}{r_{1}+r_{2}}}}-1\right)$

And $\Delta v_2$ required to transfer to the final circular orbit:

$\Delta v_2 ={\sqrt {\frac {\mu }{r_{2}}}}\left(1-{\sqrt {\frac {2r_{1}}{r_{1}+r_{2}}}}\right)$

So total,

$\Delta v =\Delta v_1+ \Delta v_2={\sqrt {\frac {\mu }{r_{1}}}}\left({\sqrt {\frac {2r_{2}}{r_{1}+r_{2}}}}-1\right)+{\sqrt {\frac {\mu }{r_{2}}}}\left(1-{\sqrt {\frac {2r_{1}}{r_{1}+r_{2}}}}\right)\\ =\sqrt {\frac {4.57*10^{14} }{1.13*10^{7}}}\left({\sqrt {\frac {2*1.48*10^{7}}{(1.13+1.48)*10^{7}}}}-1\right)+\sqrt {\frac {4.57*10^{14} }{1.48*10^{7}}}\left(1-{\sqrt {\frac {2*1.13*10^{7}}{(1.13+1.48)*10^{7}}}}\right)\\ =798.98m/s\\\approx{\color{Red} 800m/s}$