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6.1 HW1

6.1.1 Problem 1

Let us examine the accuracy of the assumption that planets orbit the Sun rather than the Sun and planet orbiting the mass center of the Sun-planet system. We’ll start with Earth:

What is the distance between the center of a spherical Sun with the radius given on your Planetary Constants sheet and the center of mass of the Sun-Earth system? Assume that the Earth is in a circular orbit about the Sun and that the ”Mean distance from the Sun” given on your Planetary Constants sheet is the distance between the mass centers of the two bodies.

Answer:

Common mass center,measured from the origin of the coordinates system is given by solving for $R$ in

$$\begin{align*} \left ( m_{1}+m_{2}\right ) R & =m_{2}r_{2}+m_{1}r_{1}\\ R & =\frac{m_{2}r_{2}+m_{1}r_{1}}{\left ( m_{1}+m_{2}\right ) } \end{align*}$$

If we now put $m_{2}$ at the center of origin, then $r_{2}=0$. Hence the above simplifies to

$$\begin{align*} \left ( m_{1}+m_{2}\right ) R & =m_{1}r_{1}\\ R & =\frac{m_{1}r_{1}}{\left ( m_{1}+m_{2}\right ) } \end{align*}$$

In our case, $m_{2}$ is the sun and $m_{1}$ is the earth, and $r_{1}$ is $AU$. Hence $$\begin{align*} R & =\frac{5.974\times 10^{24}\left [ kg\right ] \left ( 1.495978\times 10^{8}\left [ km\right ] \right ) }{\left ( 5.974\times 10^{24}\left [ kg\right ] +1.989\times 10^{30}\left [ kg\right ] \right ) }\\ & =449.32\left [ km\right ] \end{align*}$$

The above is the distance of the common center of mass of the sun-earth, measured from the center of the sun. As a percentage of the sun radius, it is $\frac{449.32}{695990}\times 100=6.4558\times 10^{-2}$ and as a percentage of the distance between the mass centers of the Sun and the Earth it is $\frac{449.32}{AU}\times 100=\frac{449.32}{1.495978\times 10^{8}}\times 100=3.0035\times 10^{-4}\%$

Summary of answers

1.

kilometers: Answer $449.319$ km

2.

percent of the Sun’s radius: Answer $0.0645\%$

3.

percent of the distance between the mass centers of the Sun and the Earth: Answer $0.000300351\%$

6.1.2 question 2

Repeat the analysis above for the most massive planet in our solar system, Jupiter.

What is the distance between the center of a spherical Sun with the radius given on your Planetary Constants sheet and the center of mass of the Sun-Jupiter system? Assume that Jupiter is in a circular orbit about the Sun and that the ”Mean distance from the Sun” given on your Planetary Constants sheet is the distance between the mass centers of the two bodies.

Answer

Now $m_{1}$ is mass of sun, but $m_{2}$ is mass of Jupiter which is $317.9$ that of the earth mass, and $r_{1}$ now is the distance from center of Jupitor to center of sun (which is the origin of the coordinates systems), which is $5.203\times AU$, hence from $$\begin{align*} \left ( m_{1}+m_{2}\right ) R & =m_{1}r_{1}\\ R & =\frac{m_{1}r_{1}}{\left ( m_{1}+m_{2}\right ) }\\ & =\frac{317.9\times \left ( 5.974\times 10^{24}\right ) \left ( 5.203\times \left ( 1.495978\times 10^{8}\right ) \right ) }{\left ( 317.9\times \left ( 5.974\times 10^{24}\right ) +1.989\times 10^{30}\right ) }\\ & =7.4248\times 10^{5}[km] \end{align*}$$

The above is the distance of the common center of mass of the sun-Jupiter, measured from the center of the sun. As a percentage of the sun radius, it is $\frac{7.4248\times 10^{5}}{695990}\times 100=$ $106.68$ $\%$ and as a percentage of the distance between the mass centers of the Sun and the Jupitor it is $\frac{7.4248\times 10^{5}}{5.203\times 1.495978\times 10^{8}}\times 100=$ $9.5391\times 10^{-4}\%$

Summary

1.

kilometers: Answer $742481$ km

2.

percent of the Sun’s radius: Answer $106.68\%$

3.

percent of the distance between the Sun and Jupiter: Answer $0.095\%$

6.1.3 question 3

A satellite is in an elliptical orbit around the Earth; at perigee its altitude is 400 km. The eccentricity of the orbit is 0.10.

6.1.3.1 part 1

What is the speed of the satellite at perigee in km/s?

answer:

$$\begin{align*} r_{p} & =r_{E}+ALT\\ & =6378+400\\ & =6778.0 \end{align*}$$

But $r_{p}=\frac{a\left ( 1-e^{2}\right ) }{1+e}$ hence $a=\frac{r_{p}\left ( 1+e\right ) }{1-e^{2}}=\frac{6778\left ( 1.1\right ) }{1-0.1^{2}}=\allowbreak 7531.1[km]$, hence $$v_{p}=\sqrt{\frac{\mu }{a}\left ( \frac{1+e}{1-e}\right ) }=\sqrt{\frac{3.986\times 10^{5}}{7531.1}\left ( \frac{1.1}{0.9}\right ) }=8.0429\left [ km/s\right ]$$

6.1.3.2 Part 2

What is the altitude of the satellite at apogee in km?

Answer

$$r_{a}=\frac{a\left ( 1-e^{2}\right ) }{1-e}=\frac{7531.1\left ( 1-0.1^{2}\right ) }{0.9}=8284.2\left [ km\right ]$$ Hence altitude $8284.2-r_{E}=8284.2-6378=1906.2\left [ km\right ]$

6.1.3.3 Part 3

What is the speed of the satellite at apogee in km/s?

Answer

$$v_{a}=\sqrt{\frac{\mu }{a}\left ( \frac{1-e}{1+e}\right ) }=\sqrt{\frac{3.986\times 10^{5}}{7531.1}\left ( \frac{0.9}{1.1}\right ) }=6.5806\left [ km/s\right ]$$

6.1.3.4 Part 4

What is the period of the orbit in hrs?

Answer

$$\begin{align*} T & =2\pi \sqrt{\frac{a^{3}}{\mu }}=2\pi \sqrt{\frac{7531.1^{3}}{3.986\times 10^{5}}}=6504.3\left [ \sec \right ] \\ & =\frac{6504.3}{60\times 60}=1.8068[hr] \end{align*}$$