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

Expert verified
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

01

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.
02

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: \[G \frac{M_{earth}m_{satellite}}{r^2} = m_{satellite}\frac{v^2}{r}\]
03

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: \[G \frac{M_{earth}}{r^2} = \frac{4 \pi^2 r^2}{T^2 r}\]
04

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: \[Altitude \approx 42,164 \text{ km} - 6,371 \text{ km} = 35,793 \text{ km}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two gears \(A\) and \(B\) are in contact. The radius of gear \(A\) is twice that of gear \(B\). (a) When \(A\) 's angular velocity is \(6.00 \mathrm{Hz}\) counterclockwise, what is \(B^{\prime}\) s angular velocity? (b) If \(A\) 's radius to the tip of the teeth is \(10.0 \mathrm{cm},\) what is the linear speed of a point on the tip of a gear tooth? What is the linear speed of a point on the tip of \(B\) 's gear tooth?
Earth's orbit around the Sun is nearly circular. The period is 1 yr \(=365.25\) d. (a) In an elapsed time of 1 d what is Earth's angular displacement? (b) What is the change in Earth's velocity, \(\Delta \overrightarrow{\mathbf{v}} ?\) (c) What is Earth's average acceleration during 1 d? (d) Compare your answer for (c) to the magnitude of Earth's instantaneous radial acceleration. Explain.
A carnival swing is fixed on the end of an 8.0 -m-long beam. If the swing and beam sweep through an angle of \(120^{\circ},\) what is the distance through which the riders move?
A velodrome is built for use in the Olympics. The radius of curvature of the surface is \(20.0 \mathrm{m}\). At what angle should the surface be banked for cyclists moving at \(18 \mathrm{m} / \mathrm{s} ?\) (Choose an angle so that no frictional force is needed to keep the cyclists in their circular path. Large banking angles are used in velodromes.)
A jogger runs counterclockwise around a path of radius \(90.0 \mathrm{m}\) at constant speed. He makes 1.00 revolution in 188.4 s. At \(t=0,\) he is heading due east. (a) What is the jogger's instantaneous velocity at $t=376.8 \mathrm{s} ?\( (b) What is his instantaneous velocity at \)t=94.2 \mathrm{s} ?$
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free