Question

Find the orbital radius of a geosynchronous satellite. Do not assume the speed found in Example \(5.9 .\) Start by writing an equation that relates the period, radius, and speed of the orbiting satellite. Then apply Newton's second law to the satellite. You will have two equations with two unknowns (the speed and radius). Eliminate the speed algebraically and solve for the radius.

Short answer

Answer: The orbital radius of a geosynchronous satellite is approximately 42,164 km. The altitude above the Earth's surface is approximately 35,793 km.

Step-by-step solution

Write down the equation relating period, radius, and orbital speed

We can find the equation that relates the period (T), radius (r), and orbital speed (v) of the satellite with the following formula: \[v = \frac{2 \pi r}{T}\] This equation comes from the fact that the circumference of the orbit is \(2 \pi r\), and the satellite travels this distance in a time period T.

Apply Newton's second law

Next, we need to apply Newton's second law. The gravitational force acting on the satellite is the centripetal force that maintains its circular orbit: \[F_{gravity} = F_{centripetal}\] The gravitational force can be represented as: \[F_{gravity} = G \frac{M_{earth}m_{satellite}}{r^2}\] where \(G\) is the gravitational constant, \(M_{earth}\) is the mass of the Earth, and \(m_{satellite}\) is the mass of the satellite. The centripetal force can be represented as: \[F_{centripetal} = m_{satellite}\frac{v^2}{r}\] Now, we can equate the two forces: \[G \frac{M_{earth}m_{satellite}}{r^2} = m_{satellite}\frac{v^2}{r}\]

Eliminate the speed

We need to eliminate the speed (v) from the equations. First, solve the equation from Step 1 for v: \[v = \frac{2 \pi r}{T}\] Now, square the equation: \[v^2 = \frac{4 \pi^2 r^2}{T^2}\] Insert this expression for \(v^2\) into the equation from Step 2: \[G \frac{M_{earth}m_{satellite}}{r^2} = m_{satellite}\frac{4 \pi^2 r^2}{T^2 r}\] We can cancel the \(m_{satellite}\) from both sides: \[G \frac{M_{earth}}{r^2} = \frac{4 \pi^2 r^2}{T^2 r}\]

Solve for the radius

Now, solve the resulting equation for the radius (r): \[r^3 = \frac{G M_{earth} T^2}{4 \pi^2}\] Taking the cube root of both sides: \[r = \sqrt[3]{\frac{G M_{earth} T^2}{4 \pi^2}}\] Since the satellite is geosynchronous, its period (T) is equal to the Earth's rotational period, 24 hours, or 86400 seconds. Plug in the known values: \[r = \sqrt[3]{\frac{(6.674 \times 10^{-11} \text{ m}^3\text{ kg}^{-1}\text{ s}^{-2})(5.972 \times 10^{24} \text{ kg})(86400\text{ s})^2}{4 \pi^2}}\] Finally, calculate the value of r: \[r \approx 42,164 \text{ km}\] This is the orbital radius of a geosynchronous satellite. Note that this value includes the radius of the Earth; to find the altitude above the Earth's surface, subtract the Earth's radius: \[Altitude \approx 42,164 \text{ km} - 6,371 \text{ km} = 35,793 \text{ km}\]