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 #65499 – Physics – Mechanics | Relativity

Question:

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:

We need to know a distance from the center of the earth to satellite (perigee distance):

$$r_p = h + R_{\text{Earth}} = 227 + 6371 = 6598\,\mathrm{km};$$

We can find semi-major axis of orbit:

$$v_p = \sqrt{GM\left(\frac{2}{r_p} - \frac{1}{a}\right)} \Rightarrow \frac{GM}{a} = \frac{2GM}{r_p} - v_p^2 \Rightarrow a = \frac{GMr_p}{2GM - r_pv_p^2} = 18724\,\mathrm{km};$$

Now we will find apogee distance and apogee speed:

$$r_a = 2a - r_p = \frac{2GMr_p}{2GM - r_pv_p^2} - r_p = 30850\,\mathrm{km};$$ $$v_a = \sqrt{GM\left(\frac{2}{r_a} - \frac{1}{a}\right)} = 2.138\,\frac{\mathrm{km}}{\mathrm{s}};$$

Answer:

$$r_a = 30850\,\mathrm{km};$$ $$v_a = 2.138\,\frac{\mathrm{km}}{\mathrm{s}};$$

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