Question

A satellite going around Earth in an elliptic orbit has a speed of 10 km s−1 at the perigee which is at a distance of 227 km from the surface of the earth. Calculate the apogee distance and its speed at that point

Answer 1

Answer on Question #65596, Physics / Mechanics | Relativity |

A satellite going around Earth in an elliptic orbit has a speed of $10\,\mathrm{km\,s^{-1}}$ at the perigee which is at a distance of $227\,\mathrm{km}$ from the surface of the earth. Calculate the apogee distance and its speed at that point.

Solution

$$\begin{array}{l} h_p = 2.27 \cdot 10^5\,\mathrm{m} \\ v_p = 10^4\,\mathrm{m/sec} \\ G = 6.67 \cdot 10^{-11}\,\mathrm{m^3\,kg^{-1}\,sec^{-2}} \text{ (universal gravitational constant)} \\ r = 6.37 \cdot 10^6\,\mathrm{m} \text{ (Earth radius)} \\ M = 5.97 \cdot 10^{24}\,\mathrm{kg} \text{ (Earth mass)} \\ v_a - ?, h_a - ? \end{array}$$

We do definitions: $r_p = r + h_p$, $r_a = r + h_a$, $a = (r_p + r_a)/2$;

From the formula

$$v = \sqrt{GM(2/r - 1/a)}$$

we find this equality

$$a = \frac{G M r_p}{2 G M - v_p^2 r_p}.$$

After the substitution of values, $a = 19\,217\,\mathrm{km}$, $r_a = 2a - r_p = 31\,837\,\mathrm{km}$, $h_a = r_a - r = 25\,467\,\mathrm{km}$.

$$\frac{v_a}{v_p} = \frac{r_p}{r_a}, \quad v_a = \frac{r_p}{r_a} \cdot v_p.$$

$v_a = 0.21 \cdot 10^4\,\mathrm{m/sec}$.

Answer: $25\,467\,\mathrm{km}$, $2.1\,\mathrm{km/sec}$.

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