Circular Motion #14

Classical Mechanics Level 4

According to Newton's law of universal gravitation, the force between two point masses $m_{1}, m_{2}$ with separation $r$ is given by

$F = \frac{ G m_{1} m_{2} }{ r^{2} }$

Here, $G$ is the universal gravitational constant and has the value $6.674 \times 10^{-11} \text{N m}^{2} \text{ kg}^{-2}$ . Use this information to solve the following problem:

A satellite orbits Earth in a circular orbit with radius of orbit $\approx 42,000 \text{ km}$ . Find the time taken (in hours) by the satellite to complete one revolution.

Assumptions and details

Mass of Earth is $6 \times 10^{ 24 } \text{ kg}$
Assume Earth and satellite to be point masses.
Ignore gravitational effects due to any other celestial body, like the Sun, Moon, other planets, etc.

This problem is part of the set - Circular Motion Practice

The answer is 23.74.

1 solution

Tom Engelsman
Dec 16, 2015

The satellite in question orbits the Earth via centripetal acceleration. Therefore, Newton's Gravitational Law can be equated as:

F = GMm/r^2 = (m*v^2)/r (i)

where M is the earth's mass and m the satellite's (in kg). Solving for the satellite's velocity in (i) gives:

v = sqrt(GM/r) = sqrt[(6.67e-11 N m^2/kg^2)(6e24 kg)/(42000e3 m)] = 3087 m/s.

Finally, the time it takes for the satellite to complete one revolution about the Earth computes:

vt/r = 2 pi radians, or t = 2 pi r/v = [2 pi*(42000e3 m) / 3087 m/s] * (1 hr/3600 s) = 23.734 hr.

Circular Motion #14

The answer is 23.74.

1 solution

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