Satellites A and B are in circular orbits, with A twice as far from Earth's center as B. How do their orbital periods compare?

Short Answer

Expert verified
The orbital period of satellite A (A) is \(2\sqrt{2}\) times longer than that of satellite B (T).

Step by step solution

01

Understand and apply Kepler's Third Law

Kepler's third law of planetary motion states that the square of the period of a planet orbiting an object (in this case Earth), is proportional to the cube of its average distance from the object. This can be expressed algebraically as:\(T^2 \propto r^3\)where \(T\) is the orbital period and \(r\) is the average distance from the center of Earth.
02

Set up the proportions

Since satellite A is twice as far from the Earth's center than satellite B, denote the distance of satellite B from Earth's center as \(r\) and that of satellite A as \(2r\). Also, let's denote the orbital period of satellite B as \'T' and that of satellite A as 'A'. According to Kepler's third law, we then have:\(T^2 = kr^3\)\(A^2 = k(2r)^3\) where \(k\) is the constant of proportionality.
03

Solve for A in terms of T

Next, we solve the equation of satellite A in terms of the equation of satellite B. We'll have to divide the second equation by the first one. This gives:\(\frac{A^2}{T^2} = \frac{k(2r)^3}{kr^3} = 8\)Taking the square root of both sides, we get:\(A = 2\sqrt{2} T.\)

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