Question

Obtain an expression for the time period of a satellite orbiting the earth. A space shuttle is in a
circular orbit at a height of 250 km from the earth’s surface, where the acceleration due to earth’s
gravity is 0.93 g. Calculate the period of its orbit. Take g = 9.8 ms−2 and the radius of the earth
R = 6.37 × 106m.

Answer 1

Obtain an expression for the time period of a satellite orbiting the earth. A space shuttle is in a circular orbit at a height of $250\mathrm{km}$ from the earth's surface, where the acceleration due to earth's gravity is $0.93\mathrm{g}$ . Calculate the period of its orbit. Take $\mathrm{g} = 9.8\mathrm{ms}^{-2}$ and the radius of the earth $\mathrm{R} = 6.37\times 10^{6}\mathrm{m}$ .

Solution.

The time period of a satellite orbiting the earth:

$$T = \frac {2 \pi (R + h)}{v};$$

$R$ - the radius of the earth;

$h$ - the height of the circular orbit;

$v$ - the speed of the space shuttle in a circular orbit.

Newton's second law:

$$m a = m g ^ {*};$$

$m$ - the mass of the space shuttle.

$g^{*}$ - the acceleration due to earth's gravity at a height of $250\mathrm{km}$ .

$$a = g ^ {*};$$ $$g ^ {*} = 0. 9 3 g;$$ $$a = 0. 9 3 g;$$

$a$ - the radial acceleration.

$$a = \frac {v ^ {2}}{(R + h)};$$ $$\frac {v ^ {2}}{(R + h)} = 0. 9 3 g;$$ $$v = \sqrt {0 . 9 3 g (R + h)};$$ $$T = \frac {2 \pi (R + h)}{\sqrt {0 . 9 3 g (R + h)}} = \frac {2 \pi}{\sqrt {0 . 9 3 g}} \cdot \frac {R + h}{\sqrt {R + h}} = 2 \pi \cdot \frac {\sqrt {R + h}}{\sqrt {0 . 9 3 g}} = 2 \pi \sqrt {\frac {R + h}{0 . 9 3 g}};$$ $$T = 2 \pi \sqrt {\frac {R + h}{0 . 9 3 g}}.$$ $$T = 2 \cdot 3. 1 4 \sqrt {\frac {6 . 3 7 \cdot 1 0 ^ {6} + 2 5 0 \cdot 1 0 ^ {3}}{0 . 9 3 \cdot 9 . 8}} = 5 3 5 2 (s).$$

Answer: $T = 5352s$.