Question

Let’s look in more detail at how a satellite is moved from one circular orbit to another. FIGURE $\text{CP13.71}$shows two circular orbits, of radii localid="1651418485730" $r_1$and localid="1651418489556" $r_2$, and an elliptical orbit that connects them. Points $1$and $2$are at the ends of the semimajor axis of the ellipse.

a. A satellite moving along the elliptical orbit has to satisfy two conservation laws. Use these two laws to prove that the velocities at points localid="1651418503699" $1$and localid="1651418499267" $2$are localid="1651418492993" $v_1^{\mathrm{\prime }}=\sqrt{\frac{2GM\left(r_2/r_1\right)}{r_1+r_2}}$and localid="1651418509687" $v_2^{\mathrm{\prime }}=\sqrt{\frac{2GM\left(r_1/r_2\right)}{r_1+r_2}}$The prime indicates that these are the velocities on the elliptical orbit. Both reduce to Equation $13.22$if localid="1651418513535" $r_1=r_{2=r}$.

b. Consider a localid="1651418519576" $1000kg$communications satellite that needs to be boosted from an orbit localid="1651418573632" $300km$above the earth to a geosynchronous orbit localid="1651418578672" $35,900km$above the earth. Find the velocity localid="1651418584351" $v_1$on the inner circular orbit and the velocity localid="1651418590277" $v=1$at the low point on the elliptical orbit that spans the two circular orbits.

c. How much work must the rocket motor do to transfer the satellite from the circular orbit to the elliptical orbit?

d. Now find the velocity localid="1651418596735" $v=2$at the high point of the elliptical orbit and the velocity v2 of the outer circular orbit.

e. How much work must the rocket motor do to transfer the satellite from the elliptical orbit to the outer circular orbit?

f. Compute the total work done and compare your answer to the result of Example localid="1651418602767" $13.6$.

Short answer

a. The velocities at points $1$and $2$are$v_1^{\text{'}}=\sqrt{\frac{2GM\left(\frac{r_2}{r_1}\right)}{r_1+r_2}}$,$v_2^{\mathrm{\prime }}=\sqrt{\frac{2GM\left(\frac{r_1}{r_2}\right)}{r_1+r_2}}$

b. The velocity $v$on the inner circular orbit and the velocity $v=1$at the low point on the elliptical orbit that spans the two circular orbits are$v_1^{\text{'}}=1.016\times {10}^4\mathrm{m}/\mathrm{s}$

c. The rocket motor do to transfer the satellite from the circular orbit to the elliptical orbit are$W\approx 2.2\times {10}^{10}\mathrm{J}$

d. The velocity $v=2$at the high point of the elliptical orbit and the velocity v2 of the outer circular orbit are$v_2^{\text{'}}=3.072\times {10}^3\mathrm{m}/\mathrm{s}$

e. The rocket motor do to transfer the satellite from the elliptical orbit to the outer circular orbit are$W=0.3432\times {10}^{10}J$

f. The total work done$r_2^{}=2.5432\times {10}^{10}\mathrm{J}$

Step-by-step solution

Angular Momentum (a)

(a) When the satellite is moving along the elliptical orbit,it's to satisfy two conservation laws:
(1) Conservation of energy
(2) Conservation of momentum
Let the rate of the satellite at points $1$ and a couple of be $v_1^{\mathrm{\prime }}$and$v_2^{\mathrm{\prime }}$, respectively.
The distance from Earth to the satellite at positions $1$ and a pair of are $r_1^{}$and $r_2^{}$, respectively.

Therefore momentum of the satellite at point $1$is $L_1=m{v}_1^{\mathrm{\prime }}{r}_1$and at point $2$is $L_2=m{v}_2^{\mathrm{\prime }}{r}_2$
From the conservation of momentum $L_1$and $L_2$

$\begin{array}{r}\Rightarrow m{v}_1^{\mathrm{\prime }}{r}_1^{\mathrm{\prime }}=m{v}_2^{\mathrm{\prime }}{r}_2\\ \end{array}$

$\Rightarrow v_1^{\mathrm{\prime }}=\left(\frac{r_2}{r_1}\right)v_2^{\mathrm{\prime }}$

Kinetic Energy

Now, the mechanical energy of the satellite at point $1$is $K_1=\frac{1}{2}m{v}_1^{\mathrm{\prime }2}$
Potential energy of the satellite at point $1$is $U_1=-\frac{GMm}{r_1}$

Therefore Total energy of the satellite at point $1$is

$\begin{array}{r}{E}_1=K_1+U_1\end{array}$

$\begin{array}{r}=\frac{1}{2}m{v}_1^{\mathrm{\prime }2}-\frac{GMm}{r_1}\\ \end{array}$

$=m\left[\frac{v_1^2}{2}-\frac{GM}{r_1^{\ast }}\right]$

Similarly, the K.E. of the satellite at point $2$ is P.E. $K_2=\frac{1}{2}m{v}_2^{\mathrm{\prime }2}$ of the satellite at point 2 is$U_2=-\frac{GMm}{r_2}$

Therefore Total energy to the satellite at point 2 is

$\begin{array}{r}{E}_2=K_2+U_2\end{array}$

$\begin{array}{r}=\frac{1}{2}m{v}_2^{\mathrm{\prime }2}-\frac{GMm}{r_2}\\ \end{array}$

$=m\left[\frac{v_2^2}{2}-\frac{GMm}{r_2}\right]$

Law of conservation of energy

Then, from the law of conservation of energy,

$\begin{array}{r}{E}_1=E_2\end{array}$

$\begin{array}{r}\Rightarrow m\left[\frac{v_2^{\mathrm{\prime }2}}{2}-\frac{GM}{r_1}\right]=m\left[\frac{v_2^{\mathrm{\prime }2}}{2}-\frac{GM}{r_2}\right]\end{array}$

$\begin{array}{r}\Rightarrow \frac{v_1^{\mathrm{\prime }2}}{2}=\frac{v_2^{\mathrm{\prime }2}}{2}-GM\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\\ \end{array}$

$\Rightarrow {\left(\frac{r_2}{r_1}\right)}^2{v}_2^{v_2^{\mathrm{\prime }2}}=v_2^{\mathrm{\prime }2}-2GM\left(\frac{r_1-r_2}{r_1{r}_2}\right)$

$\begin{array}{r}2GM\left(\frac{r_1-r_2}{r_1{r}_2}\right)=v_2^{\mathrm{\prime }2}\left(1-\frac{r_2^2}{r_1^2}\right)\\ \end{array}$

$2GM\left(\frac{r_1-r_2}{r_1{r}_2}\right)=v_2^{\mathrm{\prime }2}\frac{r_1^2-r_2^2}{r_1^2}$

Equation (a)

$\begin{array}{r}{v}_2^{\mathrm{\prime }2}=2GM\left(\frac{r_1^2}{r_1{r}_2}\right)\frac{r_1-r_2}{r_1^2-r_2^2}\end{array}$

$\begin{array}{r}=2GM\left(\frac{r_1}{r_2}\right)\frac{\left(r_1-r_2\right)}{\left(r_1-r_2\right)\left(r_1+r_2\right)}\end{array}$

$\begin{array}{r}=\frac{2GM\left(\frac{r_1}{r_2}\right)}{r_1+r_2}\\ \end{array}$

$\therefore v_2^{\mathrm{\prime }}=\sqrt{\frac{2GM\left(\frac{r_1}{r_2}\right)}{r_1+r_2}}$

Then,$v_1^{\mathrm{\prime }}=\left(\frac{r_2}{r_1}\right)\sqrt{\frac{2GM\left(\frac{r_1}{r_2}\right)}{r_1+r_2}}$

$\begin{array}{r}=\sqrt{\frac{2GM{\left(\frac{r_2}{r_1}\right)}^2\left(\frac{r_1}{r_2}\right)}{r_1+r_2}}\\ \end{array}$

$=\sqrt{\frac{2GM\left(\frac{r_2}{r_1}\right)}{r_1+r_2}}$

Circular orbit (b)

(b) the rate of the satellite at point $1$ within the circular orbit is $v_1=\sqrt{\frac{GM}{r_1}}$
Here, the mass of Earth$M=5.98\times {10}^{24}\mathrm{kg}$

$\begin{array}{r}{r}_1=h+R_e\end{array}$($R_e=$radius of earth )

$\begin{array}{r}=300\mathrm{km}+6.37\times {10}^3\mathrm{km}\end{array}$

$\begin{array}{r}=\left(6.67\times {10}^3\mathrm{km}\right)\left({10}^3\mathrm{m}/\mathrm{km}\right)\\ \end{array}$

$=6.67\times {10}^6\mathrm{m}$

$\begin{array}{r}G=6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\end{array}$

$\begin{array}{r}\therefore v_1=\sqrt{\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{\left(6.67\times {10}^6\mathrm{m}\right)}}\end{array}$

$\begin{array}{r}=\sqrt{0.598\times {10}^8{\mathrm{m}}^2/{\mathrm{s}}^2}\\ \end{array}$

$=0.77\times {10}^4\mathrm{m}/\mathrm{s}$

Equation (b)

Now, $v_1^{\mathrm{\prime }}=\sqrt{\frac{2GM\left(\frac{r_2}{r_1}\right)}{r_1+r_2}}$

Here, $r_2=35,900\mathrm{km}+6370\mathrm{km}=42270\mathrm{km}$

$=\left(42.270\mathrm{km}\right)\left({10}^3\mathrm{m}/\mathrm{km}\right)$

$=4.227\times {10}^7\mathrm{m}$

$\begin{array}{r}\therefore v_1^{\mathrm{\prime }}=\sqrt{\frac{\left(2\right)\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)\left(\frac{4.227\times {10}^7\mathrm{m}}{6.67\times {10}^6\mathrm{m}}\right)}{\left[\left(4.227\times {10}^7\mathrm{m}\right)+\left(6.67\times {10}^6\mathrm{m}\right)\right]}}\end{array}$

$\begin{array}{r}=\sqrt{\frac{50.55492\times {10}^{14}}{4.894\times {10}^7}{\mathrm{m}}^2/{\mathrm{s}}^2}\end{array}$

$\begin{array}{r}=\sqrt{10.33\times {10}^7{\mathrm{m}}^2/{\mathrm{s}}^2}\\ \end{array}$

$=1.016\times {10}^4\mathrm{m}/\mathrm{s}$

Mechanical  energy

(c) The work done is capable the change in mechanical energy

$\begin{array}{r}\therefore W=\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}_1^2\end{array}$

$\begin{array}{r}=\frac{1}{2}m\left(v_1^{\mathrm{\prime }2}-\frac{1}{2}m{v}_1^2\right)\\ \end{array}$

$=\frac{1}{2}\left(1000\mathrm{kg}\right)\left[{\left(1.016\times {10}^4\mathrm{m}/\mathrm{s}\right)}^2-{\left(0.77\times {10}^4\mathrm{m}/\mathrm{s}\right)}^2\right]$

$\begin{array}{r}=\frac{1}{2}\left(1000\mathrm{kg}\right)\left[(1.032256-0.5929)\times {10}^8{\mathrm{m}}^2/{\mathrm{s}}^2\right]\end{array}$

$\begin{array}{r}=219.678\times {10}^8\mathrm{J}\\ \end{array}$

$\approx 2.2\times {10}^{10}\mathrm{J}$

Elliptical and Circular orbit (d)

we have,

$\begin{array}{r}{v}_2^{\mathrm{\prime }}=\sqrt{\frac{2M\left(\frac{r_i}{r_2}\right)}{r_i+r_2}}\end{array}$

$\begin{array}{r}\\ =\sqrt{\left(2\right)\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)\left(\frac{6.67\times {10}^6\mathrm{m}}{4.227\times {10}^7\mathrm{m}}\right)}\\ \end{array}$

$\begin{array}{r}=\sqrt{\frac{125.878\times {10}^{12}}{4.894\times {10}^7}{\mathrm{m}}^2/{\mathrm{s}}^2}\end{array}$

$\begin{array}{r}=\sqrt{25.72\times {10}^5{\mathrm{m}}^2/{\mathrm{s}}^2}\\ \end{array}$

$=1.604\times {10}^3\mathrm{m}/\mathrm{s}$

$\begin{array}{r}{v}_2=\sqrt{\frac{GM}{r_2}}\end{array}$

$\begin{array}{r}=\sqrt{\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(5.98\times {10}^{24}\mathrm{kg}\right)}{\left(4.227\times {10}^7\mathrm{m}\right)}}\end{array}$

$\begin{array}{r}=\sqrt{9.436\times {10}^6{\mathrm{m}}^2/{\mathrm{s}}^2}\\ \end{array}$

$=3.072\times {10}^3\mathrm{m}/\mathrm{s}$

Work done (e)

(e) The work done is capable the change in mechanical energy

$\begin{array}{r}\therefore W=\frac{1}{2}m{v}_2^2\end{array}$

$\begin{array}{r}=\frac{1}{2}m{v}_2^{\mathrm{\prime }2}\\ \end{array}$

$\begin{array}{r}=\left(\frac{1}{2}\right)\left(1000\mathrm{kg}\right)\left[{\left(3.072\times {10}^3\mathrm{m}/\mathrm{s}\right)}^2-{\left(1.604\times {10}^3\mathrm{m}/\mathrm{s}\right)}^2\right]\end{array}$

$\begin{array}{r}=\left(500\mathrm{kg}\right)\left(6.864\times {10}^6{\mathrm{m}}^2/{\mathrm{s}}^2\right)\\ \end{array}$

$=3432\times {10}^6\mathrm{J}$

$\begin{array}{r}=0.3432\times {10}^{10}\mathrm{J}\end{array}$

Total work done

(f) Total work done to transfer a satellite from one lower circular orbit $\begin{array}{r}{r}_1\end{array}$ to a better circular orbit$\begin{array}{r}{r}_2\end{array}$

$\begin{array}{r}\\ =2.2\times {10}^{10}\mathrm{J}+0.3432\times {10}^{10}\mathrm{J}\\ \end{array}$

$=2.5432\times {10}^{10}\mathrm{J}$