Question

The speed of a satellite in orbit 265 km above the surface and
The period of the satellite

Answer 1

Please, find the speed of a satellite in orbit 265 km above the surface and the period of the satellite.

The second Newton's law for satellite:

$$M a = G \frac {M _ {e} M}{(R + h) ^ {2}}$$ $$a = G \frac {M _ {e}}{(R + h) ^ {2}}$$

Where $a$ is the centripetal acceleration:

$$\frac {v ^ {2}}{R + h} = G \frac {M _ {e}}{(R + h) ^ {2}}$$

So, satellite's speed:

$$v = \sqrt {G \frac {M _ {e}}{R + h}}$$

Using $g = G\frac{M_e}{R^2} \rightarrow GM_e = gR^2$

$$v = R \sqrt {\frac {g}{R + h}}$$ $$v = 6.4 * 10^{6} m \sqrt {\frac {10 m / s ^ {2}}{6 . 4 * 1 0 ^ {6} m + 0 . 2 6 5 * 1 0 ^ {6} m}} = 7. 8 4 * 1 0 ^ {3} m / s$$

Satellite's period:

$$T = \frac {2 \pi (R + h)}{v}$$ $$T = \frac {2 * 3 . 1 4 * (6 . 4 * 1 0 ^ {6} m + 0 . 2 6 5 * 1 0 ^ {6} m)}{7 . 8 4 * 1 0 ^ {3} m / s} = 5. 3 4 * 1 0 ^ {3} s$$

Answer: $v = 7.84 * 10^{3} \, \text{m/s}, T = 5.34 * 10^{3} \, \text{s}$