Answer to Question #88068 in Mechanics | Relativity for Vaishali suri

Question #88068
A satellite moves in an elliptical orbit with the Earth at one focus. At the perigee its
speed is v and its distance from the centre of the Earth is R . If the eccentricity of its
orbit is 0.5, calculate its speed at the apogee.
1
Expert's answer
2019-04-15T10:15:36-0400

Let vav_a denote the speed of the satellite at the apogee to be found, and let its distance from the center of the Earth at the apogee be RaR_a. The conservation of energy of the satellite gives the equation

mva22GmMRa=mv22GmMR,\frac{m v_a^2}{2} - \frac{G m M}{R_a} = \frac{m v^2}{2} - \frac{G m M}{R} \, ,

where mm and MM are the masses of the satellite and of the Earth, respectively, and GG is Newton's gravitational constant. Solving this equation with respect to vav_a, we have

va=v2GMR(1RRa).v_a = \sqrt{ v^2 - \frac{G M}{R} \left( 1 - \frac{R}{R_a} \right) } \, .

The ratio of the distances in the perigee and apogee is related to the eccentricity ϵ\epsilon as

RRa=1ϵ1+ϵ.\frac{R}{R_a} = \frac{1 - \epsilon}{1 + \epsilon} \, .

Since ϵ=1/2\epsilon = 1/2, we obtain R/Ra=1/3R / R_a = 1/3 and

va=v22GM3R.v_a = \sqrt{ v^2 - \frac{2 G M}{3 R} } \, .



Answer: va=v22GM3R.v_a = \sqrt{ v^2 - \frac{2 G M}{3 R} } \, .


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