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Find the radius of the orbit of a synchronous satellite that circles the Earth. (A synchronous satellite goes around the Earth once every \(24 \mathrm{~h}\), so that its position appears stationary with respect to a ground station.) The simplest way to find the answer and give your results is by expressing all distances in terms of the Earth's radius \(R_{e}\)

Short Answer

Expert verified
The radius of a synchronous satellite orbit is approximately 6.6 \(R_e\), where \(R_e\) is the Earth's radius.

Step by step solution

01

Understand the Synchronous Orbit

A synchronous satellite completes one revolution around the Earth in 24 hours. This means the period \(T\) of the satellite is equal to 24 hours, or 86400 seconds. The satellite remains above the same point on Earth as it has the same rotational period as the Earth.
02

Use Kepler's Third Law

Kepler's Third Law relates the period of orbit \(T\) to the radius of the orbit \(r\) and mass of Earth \(M_e\). The formula is \(T^2 = \frac{4\pi^2}{GM_e} r^3\), where \(G\) is the gravitational constant and \(M_e\) is the mass of the Earth.
03

Rearrange for Radius

Rearrange Kepler's Third Law to solve for the radius of the orbit \(r\): \[ r^3 = \frac{GMT^2}{4\pi^2} \]
04

Substitute Known Values

Substitute \(T = 86400\) seconds, \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\), and \(M_e = 5.972 \times 10^{24} \text{ kg}\) into the formula: \[ r^3 = \frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times (86400)^2}{4\pi^2} \]
05

Calculate Radius Cube

Calculate \(r^3\) using the substituted values. Compute \(\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times (86400)^2}{4\pi^2}\) to find the value of \(r^3\).
06

Solve for Radius \(r\)

Take the cube root of \(r^3\) to find \(r\). This gives \(r = (r^3)^{1/3}\).
07

Express Radius in Terms of Earth's Radius

Convert \(r\) from meters to Earth’s radius by dividing \(r\) by the Earth’s radius \(R_e = 6.371 \times 10^6\) meters, so that \(\frac{r}{R_e}\) gives the radius in terms of Earth's radius \(R_e\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kepler's Third Law
Kepler's Third Law is a fundamental principle in astronomy that relates the time a celestial body takes to orbit (its period) to the size of its orbit. For planets around the sun, it states that the square of the period of orbit (T^2) is proportional to the cube of the semi-major axis (orbital radius) of its orbit (\frac{4\pi^2}{GM_e} r^3.). This same principle can be adapted to satellites orbiting Earth. Here, the formula \(T^2 = \frac{4\pi^2}{GM_e} r^3\) is used. In this case, \(T\) represents the period of the satellite (24 hours for synchronous satellites), while \(r\) is the orbital radius, and \(G\) is the gravitational constant. Understanding Kepler's Third Law helps explain why satellites that orbit further out take longer to complete their orbits.
Orbital Radius Calculation
Calculating the orbital radius of a satellite involves some interesting mathematics and applications of physics principles. For a synchronous satellite, which orbits in sync with Earth's rotation, you first set the orbital period to 24 hours, or 86400 seconds, since 1 hour equals 3600 seconds. Using the adapted version of Kepler's Third Law for Earth, the formula is \(r^3 = \frac{GMT^2}{4\pi^2}\), where \(G\) is the gravitational constant and \(M_e\) is Earth's mass.
  • First, substitute the known values into the formula: \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\) and \(M_e = 5.972 \times 10^{24} \text{ kg}\).
  • This step involves some number crunching to calculate \(r^3\).
  • Finally, to solve for \(r\), take the cube root of \(r^3\). This gives the orbital radius in meters.
Remembering these steps might help you tackle similar problems in the future.
Earth's Radius
The Earth's radius is a critical component when discussing satellite orbits, especially in exercises like deriving the radius of a synchronous satellite in terms of Earth's radius \(R_e\). Earth's average radius is approximately \(6.371 \times 10^6\) meters. When we express the satellite's orbital radius in terms of Earth's radius, it simplifies comparisons and calculations, making it easy to grasp the scale of distances involved.
To convert the calculated orbital radius into a more comprehensive format, you simply divide the orbital radius by the Earth's radius \(R_e\). This ratio helps us better understand the satellite's position relative to our planet. It's important for applications where knowing the exact distance in relation to Earth's radius provides meaningful insight for practical uses, like satellite communications or scientific observations.

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Most popular questions from this chapter

A wheel of radius \(R\) rolls along the ground with velocity \(V .\) A pebble is carefully released on top of the wheel so that it is instantaneously at rest on the moving wheel. (a) Show that the pebble will immediately fly off the wheel if \(V>\sqrt{R g}\) (b) Show that in the case where \(V<\sqrt{R g}\), and the coefficient of friction is \(\mu=1\), the pebble starts to slide when it has rotated through an angle given by \(\theta=\arccos \left[(1 / \sqrt{2})\left(V^{2} / R g\right)\right]-\pi / 4\).

A device called a capstan is used aboard ships in order to control a rope which is under great tension. The rope is wrapped around a fixed drum, usually for several turns (the drawing shows about a three-quarter turn). The load on the rope pulls it with a force \(T_{A}\), and the sailor holds it with a much smaller force \(T_{B}\). Show that \(T_{B}=T_{A} e^{-\mu \theta}\), where \(\mu\) is the coefficient of friction and \(\theta\) is the total angle subtended by the rope on the drum.

A uniform rope of weight \(W\) hangs between two trees. The ends of the rope are the same height, and they each make angle \(\theta\) with the trees. Find (a) The tension at either end of the rope. (b) The tension in the middle of the rope.

A car is driven on a large revolving platform which rotates with constant angular speed \(\omega\). At \(t=0\) a driver leaves the origin and follows a line painted radially outward on the platform with constant speed \(v_{0}\). The total weight of the car is \(W\), and the coefficient of friction between the car and stage is \(\mu\). (a) Find the acceleration of the car as a function of time using polar coordinates. Draw a clear vector diagram showing the components of acceleration at some time \(t>0\). (b) Find the time at which the car just starts to skid. (c) Find the direction of the friction force with respect to the instantaneous position vector \(\mathbf{r}\) just before the car starts to skid. Show your result on a clear diagram.

A small bead of mass \(m\) is free to slide on a thin rod. The rod rotates in a plane about one end at constant angular velocity \(\omega\). Show that the motion is given by \(r=A e^{-\gamma^{\prime}}+B e^{+\gamma^{\prime}}\), where \(\gamma\) is a constant which you must find and \(A\) and \(B\) are arbitrary constants. Neglect gravity. Show that for a particular choice of initial conditions [that is, \(r(t=0)\) and \(v(t=0)]\) it is possible to obtain a solution such that \(r\) decreases continually in time, but that for any other choice \(r\) will eventually increase. (Exclude cases where the bead hits the origin.)

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