Question #274928

An earth satellite has a speed of 28070 km/hr when it is at its perigee of 220 km above Earth's surface. Find the apogee distance its speed at apogee, and its period of revolution.


1
Expert's answer
2021-12-03T12:44:29-0500

From the speed at the perigee, find the semi-major axis of the satellite's orbit:



vper=GM(2rper1a), a=GMrper2GMvper2rper=GM(RE+hper)2GMvper2(RE+hper)=6636 km.v_\text{per}=\sqrt{GM\bigg(\frac 2r_\text{per}-\frac1a\bigg)},\\\space\\ a=\frac{GMr_\text{per}}{2GM-v_\text{per}^2r_\text{per}}=\frac{GM(R_E+h_\text{per})}{2GM-v_\text{per}^2(R_E+h_\text{per})}=6636\text{ km}.



Find the apogee distance:


rap=2arper=6677 km.r_\text{ap}=2a-r_\text{per}=6677\text{ km}.



From the apogee distance, find the speed at the apogee:

vap=GM(2rap1a)=7.7 km.v_\text{ap}=\sqrt{GM\bigg(\frac 2r_\text{ap}-\frac1a\bigg)}=7.7\text{ km}.

Apply Kepler's third law to find the period of revolution:


TMoon2T2=aMoon3a3, T=TMoona3a Moon3=0.062 days,\frac{T_\text{Moon}^2}{T^2}=\frac{a^3_\text{Moon}}{a^3},\\\space\\ T=T_\text{Moon}\sqrt{\frac{a^3}{a^3_\text{ Moon}}}=0.062\text{ days},

or around 1 h 30 min.


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