Answer to Question #274928 in Classical Mechanics for zaw myo

Question

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.

Answer 1

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

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

Find the apogee distance:

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

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

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

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

$$\frac{T_\text{Moon}^2}{T^2}=\frac{a^3_\text{Moon}}{a^3},\\\space\\ T=T_\text{Moon}\sqrt{\frac{a^3}{a^3_\text{ Moon}}}=0.062\text{ days},$$

or around 1 h 30 min.