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Chapter 17: Problem 32

A satellite orbits the Earth \(200 \mathrm{~km}\) above the surface. What speed, in \(\mathrm{m} / \mathrm{s}\), is necessary for a circular orbit? The radius of the Earth is \(6400 \mathrm{~km}\) and \(\mathrm{g}=9.2 \mathrm{~m} / \mathrm{s}^{2}\) a) 7800 c) 6600 b) 7200 d) 6000

Short Answer

Expert verified
(a) 7800 m/s

Step by step solution

01

Write down the known values

The known values are: - Distance from Earth's surface to satellite (height): \(h = 200 \mathrm{~km}\) - Earth's radius: \(R = 6400 \mathrm{~km}\) - Gravitational acceleration: \(g = 9.2 \mathrm{~m} / \mathrm{s}^{2}\) Note that we need to convert the given heights from km to meters.
02

Convert km to meters

To convert km to meters, we multiply by 1000: - \(h = 200 \mathrm{km} \times 1000 = 200,000 \mathrm{m}\) - \(R = 6400 \mathrm{km} \times 1000 = 6,400,000 \mathrm{m}\)
03

Calculate the satellite's distance from the Earth's center

The distance from the satellite to the Earth's center is the sum of the Earth's radius and the distance from the Earth's surface to the satellite: $$ r = R + h = 6,400,000 \mathrm{m} + 200,000 \mathrm{m} = 6,600,000 \mathrm{m} $$
04

Equate gravitational force and centripetal force

The gravitational force acting on the satellite is given by: $$ F_g = \frac{GMm}{r^2} $$ where \(G\) is the gravitational constant, \(M\) is the Earth's mass, and \(m\) is the mass of the satellite. The centripetal force acting on the satellite is given by: $$ F_c = \frac{mv^2}{r} $$ where \(v\) is the speed of the satellite. Since the satellite is in a circular orbit, these two forces must be equal: $$ F_g = F_c $$
05

Solve for the satellite's speed

Having equated the gravitational force and centripetal force, we can eliminate satellite's mass \(m\): $$ \frac{GMm}{r^2} = \frac{mv^2}{r} $$ Divide both sides of the equation by \(m\) and \(r\). Then, we get: $$ \frac{GM}{r} = v^2 $$ Now, isolate the speed \(v\): $$ v = \sqrt{\frac{GM}{r}} $$ We can rewrite the gravitational acceleration as \(g = \frac{GM}{R^2}\): $$ \frac{gR^2}{r} = v^2 $$ Substitute the known values for gravitational acceleration, Earth's radius, and satellite's distance from the Earth's center into the equation: $$ v = \sqrt{\frac{9.2 \frac{\mathrm{m}}{\mathrm{s}^2}(6,400,000 \mathrm{m})^2}{6,600,000 \mathrm{m}}} $$
06

Calculate the satellite's speed

Plug the values into the equation and solve for \(v\): $$ v = \sqrt{\frac{9.2 (6,400,000)^2}{6,600,000}} \approx 7800 \mathrm{m} / \mathrm{s} $$ So, the speed necessary for a satellite to maintain a circular orbit at a distance of 200 km above the Earth's surface is approximately 7800 m/s. The correct answer is (a) 7800.

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