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 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 × 10⁶ m.

Solution:

Let T- the time period of a satellite orbiting the earth.

h = 250 km = 250 000 m = 2.5*10⁵ m – height from earth's surface

R = 6.67 10⁶ m – the radius of the earth

$$v - \text{velocity of a satellite}$$ $$T = \frac{2\pi(R + h)}{v}$$

So we find the velocity of satellite.

If m – mass of satellite and M-mass of Earth and γ – gravitational constant

$$F = \gamma m M / ((R + h)^2)$$

In other hand $F = ma = m \frac{v^2}{R + h} - v = \sqrt{\gamma \frac{M}{R + h}} \cdot \sqrt{\frac{\gamma M}{R^2} \frac{R^2}{R + h}} = \sqrt{g \frac{R^2}{R + h}}$

So

$$T = \frac{2\pi(R + h)}{v} = \frac{2\pi(R + h)}{\sqrt{g \frac{R^2}{R + h}}} = \frac{2\pi(R + h)^{3/2}}{R \sqrt{g}} = \frac{2 \cdot 3.14 \cdot (6.67 \cdot 10^6 + 2.5 \cdot 10^5)^{\frac{3}{2}}}{6.67 \cdot 10^6 \sqrt{9.81}} = 5472 \text{ seconds}$$ $$= 1.52 \text{ hour}$$

Answer: 1.52 hours