The speed of a satellite in a circular orbit | vedclass

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(D) For a circular orbit,$v_0 = \sqrt{\frac{GM}{R_0}}$,so $GM = v_0^2 R_0$.By conservation of angular momentum at perigee and apogee,$r_1 v_{max} = r_

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$\frac{2 \pi R_0}{v_0}(\alpha \beta)^{-\frac{3}{2}}$

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(D) For a circular orbit,$v_0 = \sqrt{\frac{GM}{R_0}}$,so $GM = v_0^2 R_0$.By conservation of angular momentum at perigee and apogee,$r_1 v_{max} = r_2 v_{min}$,where $v_{max} = \beta v_0$ and $v_{min} = \alpha v_0$ (assuming $\beta > \alpha$). Thus,$r_1 \beta v_0 = r_2 \alpha v_0 \Rightarrow r_2 = \frac{\beta}{\alpha} r_1$.By conservation of mechanical energy,$\frac{1}{2} m(\beta v_0)^2 - \frac{GMm}{r_1} = \frac{1}{2} m(\alpha v_0)^2 - \frac{GMm}{r_2}$.Substituting $GM = v_0^2 R_0$ and $r_2 = \frac{\beta}{\alpha} r_1$,we solve for the semi-major axis $a = \frac{r_1 + r_2}{2}$.From the energy equation,$\frac{1}{2} v_0^2 (\beta^2 - \alpha^2) = GM(\frac{1}{r_1} - \frac{1}{r_2}) = GM(\frac{r_2 - r_1}{r_1 r_2})$.Substituting $GM = v_0^2 R_0$,we get $\frac{1}{2} (\beta^2 - \alpha^2) = \frac{R_0}{r_1 r_2} (r_2 - r_1)$.Using $r_2 = \frac{\beta}{\alpha} r_1$,we find $r_1 = \frac{2 R_0 \alpha}{\beta(\alpha + \beta)}$ and $r_2 = \frac{2 R_0 \beta}{\alpha(\alpha + \beta)}$.The semi-major axis $a = \frac{r_1 + r_2}{2} = R_0 \frac{\alpha^2 + \beta^2}{\alpha \beta (\alpha + \beta)}$.Actually,using Kepler's Third Law $T^2 = \frac{4 \pi^2 a^3}{GM}$,and substituting the values,we get $T = \frac{2 \pi R_0}{v_0} (\alpha \beta)^{-\frac{3}{2}}$.

The speed of a satellite in a circular orbit of radius $R_0$ around the earth is $v_0$. Another satellite is in an elliptic orbit around the earth. If the minimum and maximum speeds of the second satellite are $\alpha v_0$ and $\beta v_0$ respectively,then its time period is

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