Question

A spy satellite designed to peer closely at a particular house every day at noon has a 24-h period, and a perigee of $100 \mathrm{~km}$ directly above the house. What is the altitude of the satellite at apogee? (Earth's radius is $6400 \mathrm{~km}$.)

   A spy satellite designed to peer closely at a particular house every day at noon has a 24-h period, and a perigee of $100 \mathrm{~km}$ directly above the house. What is the altitude of the satellite at apogee? (Earth's radius is $6400 \mathrm{~km}$.)
 
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Modern Classical Mechanics
Modern Classical Mechanics
T. M. Helliwell, V.… 1st Edition
Chapter 7, Problem 9 ↓
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A spy satellite designed to peer closely at a particular house every day at noon has a 24-h period, and a perigee of $100 \mathrm{~km}$ directly above the house. What is the altitude of the satellite at apogee? (Earth's radius is $6400 \mathrm{~km}$.)
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Key Concepts

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Kepler's Third Law
Kepler's Third Law relates the orbital period of a body to the size of its orbit, specifically stating that the square of the orbital period is proportional to the cube of the semi?major axis. This law is essential in calculating orbital distances and determining the size of an orbit when the period is known.
Elliptical Orbits
Elliptical orbits are the paths followed by bodies under central gravitational forces, where the orbit is shaped as an ellipse rather than a circle. This concept introduces key definitions such as perigee (the closest point to the central body) and apogee (the farthest point), which are fundamental in understanding variations in orbital altitude.
Perigee and Apogee
Perigee and apogee are critical points in an elliptical orbit, representing the minimum and maximum distances from the orbiting body to the central body, respectively. Knowing one of these, along with the orbital period, allows for the determination of the other by relating them through the orbital geometry.
Semi-Major Axis and Eccentricity
The semi-major axis is half of the longest diameter of an ellipse and serves as a measure of the overall size of the orbit, while the eccentricity quantifies its deviation from a perfect circle. Together, these parameters define the shape and energy of the orbit, and are instrumental in calculating specific orbital characteristics such as distance at perigee and apogee.

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Transcript

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00:01 So for this problem, given the orbital period of a satellite going around the earth, we want to determine how far off the earth's surface it's sitting.
00:14 So in order to do this, we need an equation that relates the period of an orbit to the radius of that orbit.
00:24 And that's going to be kepler's law.
00:27 Kepler's third law, specifically this right here.
00:35 So this r in kepler's law here is the distance from the center of the object we're orbiting all the way out to the object that's orbiting.
00:47 So you can see from our drawing here that that's actually going to be the sum of r sub e, the radius of the earth, and r sub s, the distance from the surface to the orbiting object.
01:00 So what we should be able to do here is just substitute in for r and solve for r sub s, which is the distance that the satellite is sitting off the surface of the earth.
01:16 Everything else should be known.
01:18 We'll have to look up the mass of the earth and the radius of the earth, but these are all just constants.
01:23 So let's go ahead and substitute in 4 pi squared over gm's e times r sub e plus r sub s cubed.
01:51 I'm going to multiply both sides by gm over 4 pi squared.
01:58 So we should have g m sub e t squared over 4 pi squared equals r sub e plus r sub s cubed and then we'll take the cube root of both sides and subtract r sub e so what we left with r sub s equals all this stuff g m sub e t squared over 4 pi squared the 1 third power since we took the cube root minus r sub e okay so now i all we have to do is substitute in all our values here.
02:55 So remember the universal gravitational constant is about 6 .67 times 10 to the minus 11 meters cubed over kilogram second square...
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