Question

A satellite going around the earth in an elliptical orbit has a speed of 10 km/s at the perigee which is at a distance of 227 km from the surface of the earth. Calculate the apogee distance &its speed at that point.

Answer 1

Answer on Question #66743, Physics / Mechanics | Relativity

Question:

A satellite going around the earth in an elliptical orbit has a speed of $10\mathrm{km / s}$ at the perigee which is at a distance of $227\mathrm{km}$ from the surface of the earth. Calculate the apogee distance & its speed at that point.

Solution:

The satellite's speed at the perigee may be calculated according to this formula:

$v_{p} = \sqrt{\frac{GM}{r_{p}}(1 + e)}$ , where $G$ — gravitational constant (6.674×10 $^{-11}$ m $^3$ kg $^{-1}$ s $^{-2}$ );

M - the Earth's mass (5.972×10 $^{24}$ kg);

$r_p$ - perigee distance from the centre of the earth;

$e$ - eccentricity of the orbit.

Then $e = \frac{v_p^2 r_p}{GM} - 1 = \frac{10000^2 \cdot (6371000 + 227000)}{6.674 \times 10^{-11} \cdot 5.972 \times 10^{24}} - 1 = 0.655$

In apogee the satellite's speed:

$v_{a} = \sqrt{\frac{GM}{r_{a}}(1 - e)}$ , where $r_a$ - apogee distance from the centre of the earth.

For elliptical orbit the distances $r_p$ and $r_a$ are related as $\frac{r_a}{r_p} = \frac{1 + e}{1 - e}$ , and then

$r_a = r_p \frac{1 + e}{1 - e} = (6371000 + 227000) \frac{1 + 0.655}{1 - 0.655} = 31651275 \, m$ .

The apogee distance from the surface of the earth

$d_{a} = 31651275 - 6371000 = 25280275 \, m \cong 25280 \, km.$

$v_{a} = \sqrt{\frac{6.674\times 10^{-11}\cdot 5.972\times 10^{24}}{31651275}(1 - 0.655)} = 2084\frac{m}{s}\cong 2.1km / s$

Answer:

25280 km

2.1 km/s

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