Question

Obtain an expression for the time period of a satellite orbiting the earth. At what altitude
should a satellite be placed for its orbit to be geosynchronous?

Answer 1

Obtain an expression for the time period of a satellite orbiting the earth. At what altitude should a satellite be placed for its orbit to be geosynchronous?

Second Newton's law for this case:

$$G \frac {M _ {E} m}{(R _ {E} + h) ^ {2}} = m \frac {v ^ {2}}{R _ {E} + h}$$ $$G \frac {M _ {E}}{R _ {E} + h} = v ^ {2}$$ $$v = \sqrt {G \frac {M _ {E}}{R _ {E} + h}} = \sqrt {\frac {g R _ {E} ^ {2}}{R _ {E} + h}} = R _ {E} \sqrt {\frac {g}{R _ {E} + h}}$$

One revolution will be made during time (period):

$$T = \frac {2 \pi (R _ {E} + h)}{R _ {E} \sqrt {\frac {g}{R _ {E} + h}}} = \frac {2 \pi (R _ {E} + h) ^ {3 / 2}}{R _ {E} \sqrt {g}}$$

Assuming that the Earth period is 24h:

$$\frac {2 \pi (R _ {E} + h) ^ {3 / 2}}{R _ {E} \sqrt {g}} = 24 * 3600s$$ $$h = \left(\frac {86400s * R _ {E} \sqrt {g}}{2 \pi}\right) ^ {2 / 3} - R _ {E}$$ $$h = \left(\frac {86400s * 6400000m * \sqrt {9.8m / s ^ {2}}}{2 * 3.14}\right) ^ {2 / 3} - 6400000m = 3.6 * 10 ^ {7} m = 3.6 * 10 ^ {4} km$$

Answer: $T = \frac{2\pi(R_E + h)^{3/2}}{R_E\sqrt{g}}, h = 3.6 * 10^4 km$