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Chapter 6: Problem 770

If the height of a satellite from the earth is negligible in comparison of the radius of the earth \(\mathrm{R}\), the orbital velocity of the satellite is (A) \(\mathrm{gR}\) (B) \((\mathrm{gR} / 2)\) (C) \(\sqrt{(g} / \mathrm{R})\) (D) \(\sqrt{(g R)}\)

Short Answer

Expert verified
The orbital velocity of the satellite can be found by equating the gravitational force and the required centripetal force to keep it in orbit. This results in the equation \(v^2 = gR\). Solving for orbital velocity, we get \(v = \sqrt{gR}\), which corresponds to answer (D).

Step by step solution

01

Identify centripetal force and gravitational force

Let's denote the mass of the satellite as m, the Earth's gravitational constant as g, and the Earth's radius as R. The centripetal force acting on the satellite is given by: \[F_c = \frac{mv^2}{R}\] Where v is the orbital velocity and R is the distance between the Earth's center and the satellite (which is still the Earth's radius). The gravitational force acting on the satellite is given by: \[F_g = mg\]
02

Equate the two forces

In order for the satellite to stay in orbit, the gravitational force must be equal to the centripetal force. Therefore: \[\frac{mv^2}{R} = mg\]
03

Solve for orbital velocity, v

Simplify the equation and solve for v. First, divide both sides by m: \[\frac{v^2}{R} = g\] Now, multiply by R to isolate v^2: \[v^2 = gR\] Finally, take the square root of both sides to get v: \[v = \sqrt{gR}\] Therefore, the orbital velocity of the satellite is given by answer (D): \[v = \sqrt{(g R)}\]

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