To teach you how to find the parameters characterizing an object in a circular o

Question

To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M.

For all parts of this problem, where appropriate, use G for the universal gravitational constant.

Part G

The quantities v, K, U, and L all represent physical quantities characterizing the orbit that depend on radius R. Indicate the exponent (power) of the radial dependence of the absolute value of each.

Express your answer as a comma-separated list of exponents corresponding to v, K, U, and L, in that order. For example, -1,-1/2,-0.5,-3/2 would mean vR1, KR1/2, and so forth.

Exercise 8.26

A rocket is launched vertically upward from Earth's surface at a speed of 5.5 km/s .

Part A

What is its maximum altitude?

Exercise 8.30

The escape speed from a planet of mass 3.5×1024 kg is 6.8 km/s .

Part A

What is the planet's radius?

Problem 8.39

During the Apollo Moon landings, one astronaut remained with the command module in lunar orbit, about 130 km above the surface. For half of each orbit, this astronaut was completely cut off from the rest of humanity, as the spacecraft rounded the far side of the Moon.

Part A

How long did this period last?

Explanation / Answer

8.26) Now potential at earths surface = -GMm/r = -GMm/r^2 *r = -mgr

and potential at maximum altitude = GMm/(r+x)= -GMm/r^2 *r^2/(r+x) =m gr^2/(r+x) where x is the height above the earths surface.

I have made these alterations to avoid having to know the mass of the earth. We can measure g however.
r = 6.4 * 10 ^ 6 m, g = 9.8

so 1/2 m v^2 = final - initial = -mgr^2/(r+x) + mgr
1/2 v^2 = - g* r^2 (r+x)+ gr
1/2 v^2/gr = -r/(r+x) +1
1/2 v^2 / gr -1 = -r/(r+x)
1- 1/2 v^2/gr= r/(r+x)
substitute g and r

1-1/2 * (5500)^2/9.8/(6.4*10^6)= r/(r+x)
0.7588= r/(r+x)
0.7588 (r+x) =r
0.7588x = (1-0.7588)r
x = r*(1-0.7588)/ 0.7588
=2.03* 10 ^6 m
so that is the maximum height attained by this rocket.

8.30)

v (escape) = (2 r g)
r = radius
g = acceleration due to gravity of that planet
g = GM/r^2
g*r = GM/r
v (escape) = (2 GM/r)
v^2 (escape) = 2 GM/r

r = 2 GM / v^2 (escape)
r = radius = 2*6.67*10^-11*3.5*10^24 / [6.8*1000]^2
r = 1.009 *10^7 meter

8.39)

M = 7.36 * 10 ^ 22 kg
G = 6.673 * 10 ^ -11
rM = 1737400 m
ro = (130000 + rM) m
T^2 = ro^3 * 4 * pi^2/(GM)
T = Sqrt(ro ^ 3 * 4 * pi ^ 2 / (G * M))
= 7234.96 s

Half orbit period
= T/2 = 7234.96 /2 = 3617.48 s

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