Chapter 13.2, Problem 13.116P

A spacecraft of mass m describes a circular orbit of radius r₁ around the earth. (a) Show that the additional energy ΔE that must be imparted to the spacecraft to transfer it to a circular orbit of larger radius r₂ is

$\Delta E=\frac{GMm(r_2-r_1)}{2r_2{r}_1}$

where M is the mass of the earth. (b) Further show that if the transfer from one circular orbit to the other is executed by placing the spacecraft on a transitional semielliptic path AB, the amounts of energy ΔE_A and ΔE_B which must be imparted at A and B are, respectively, proportional to r₂ and r₁:

$\begin{array}{ll}\Delta E_A=\frac{r_2}{r_2+r_1}\Delta E\hfill & \Delta E_B=\frac{r_1}{r_1+r_2}\Delta E\hfill \end{array}$

Fig. P13.116

To determine

Show that additional energy $\left(\Delta E\right)$ that must be imparted to the spacecraft to transfer it to a circular orbit of larger radius $\left(r_2\right)$ is $\Delta E=\frac{GMm\left(r_2-r_1\right)}{2r_1{r}_2}$.

Answer to Problem 13.116P

The additional energy $\left(\Delta E\right)$ that must be imparted to the spacecraft to transfer it to a circular orbit of larger radius $\left(r_2\right)$ is $\underset{\_}{\Delta E=\frac{GMm\left(r_2-r_1\right)}{2r_1{r}_2}}$ and hence it is proved.

Explanation of Solution

Given information:

The minimum distance between the center of the earth to the point A is $r_1$.

The maximum distance between the center of the earth to the point B is $r_2$.

The mass of the earth is M.

Calculation:

Show the figure with the force acting as in Figure (1).

The expression for the normal acceleration $\left(a_n\right)$ as follows:

$a_n=\frac{v^2}{r}$

The expression for calculating the geocentric force acting on the spacecraft when it is on the surface of earth (F) as follows:

$F=\frac{GMm}{r^2}$

Here, G is the universal gravitational constant and M is the mass of the earth.

Calculate the velocity of the circular orbit (v) by considering the force equilibrium by taking Newton’s second law using the relation:

$F=m{a}_n$

Substitute $\frac{GMm}{r^2}$ for F and $\frac{v^2}{r}$ for $a_n$.

$\begin{array}{l}F=m{a}_n\\ \frac{GMm}{r^2}=\frac{m{v}^2}{r}\\ \frac{GM}{r^2}=\frac{v^2}{r}\\ \frac{GM}{r}=v^2\\ v=\sqrt{\frac{GM}{r}}\end{array}$ (1)

The expression for the kinetic energy in the circular orbit (T) as follows;

$T=\frac{1}{2}m{v}^2$

The expression for the potential energy in the circular orbit (V) as follows;

$V=-\frac{GMm}{r}$

Calculate the energy required (E) for the spacecraft using the relation:

$E=T+V$

Substitute $\frac{1}{2}m{v}^2$ for T and $-\frac{GMm}{r}$ for V.

$E=\frac{1}{2}m{v}^2-\frac{GMm}{r}$

Substitute $\frac{GM}{r}$ for $v^2$.

$\begin{array}{c}E=\frac{1}{2}m\frac{GM}{r}-\frac{GMm}{r}\\ =\frac{GMm-2GMm}{2r}\\ =-\frac{1}{2}\frac{GMm}{r}\end{array}$

The expression for the energy required for the circular orbit of radius $\left(r_1\right)$ as follows:

$E_1=-\frac{1}{2}\frac{GMm}{r_1}$

The expression for the energy required for the circular orbit of radius $\left(r_2\right)$ as follows:

$E_2=-\frac{1}{2}\frac{GMm}{r_2}$

Calculate the addition energy imparted to the spacecraft to transfer it to circular orbit $\left(\Delta E\right)$ using the relation:

$\Delta E=E_1-E_2$

Substitute $-\frac{1}{2}\frac{GMm}{r_1}$ for $E_1$ and $-\frac{1}{2}\frac{GMm}{r_2}$ for $E_2$.

$\begin{array}{c}\Delta E=-\frac{1}{2}\frac{GMm}{r_1}-\left(-\frac{1}{2}\frac{GMm}{r_2}\right)\\ =-\frac{1}{2}\frac{GMm}{r_1}+\frac{1}{2}\frac{GMm}{r_2}\\ =\frac{GMm{r}_2-GMm{r}_1}{2r_1{r}_2}\\ =\frac{GMm\left(r_2-r_1\right)}{2r_1{r}_2}\end{array}$

Therefore, the additional energy $\left(\Delta E\right)$ that must be imparted to the spacecraft to transfer it to a circular orbit of larger radius $\left(r_2\right)$ is $\underset{\_}{\Delta E=\frac{GMm\left(r_2-r_1\right)}{2r_1{r}_2}}$ and hence it is proved.

To determine

Show the transfer from one circular orbit to the other is executed by placing the spacecraft on transitional semi elliptic path AB, the amounts of energy $\left(\Delta E_A\right)$ and $\left(\Delta E_B\right)$ imparted at A and B are proportional to $\left(r_2\right)$ and $\left(r_1\right)$ is defined by,

$\Delta E_A=\frac{r_2}{r_1+r_2}\Delta E$ and $\Delta E_B=\frac{r_1}{r_1+r_2}\Delta E$.

Answer to Problem 13.116P

The amount of energy imparted at A $\left(\Delta E_A\right)$ and B $\left(\Delta E_B\right)$ are $\underset{\_}{\Delta E_A=\frac{r_2}{r_1+r_2}\Delta E}$ and $\underset{\_}{\Delta E_B=\frac{r_1}{r_1+r_2}\Delta E}$ respectively and hence it is proved.

Explanation of Solution

Given information:

The minimum distance between the center of the earth to the point A is $r_1$.

The maximum distance between the center of the earth to the point B is $r_2$.

The mass of the earth is M.

Calculation:

Consider the circular orbit of radius $\left(r_1\right)$.

The expression for the velocity of the circular orbit $\left(v_{circ}\right)$ as follows:

$v_{circ}^2=\frac{GM}{r_1}$

Calculate the kinetic energy at the circular orbit $\left(T_{circ}\right)$ using the formula:

$T_{circ}=\frac{1}{2}m{v}_{circ}^2$

Substitute $\frac{GM}{r_1}$ for $v_{circ}^2$.

$\begin{array}{c}{T}_{circ}=\frac{1}{2}m\left(\frac{GM}{r_1}\right)\\ =\frac{GMm}{2r_1}\end{array}$

Consider that the after the spacecraft engines are fired and it is placed on a semi-elliptic path AB.

The expression or the principle of conservation of angular momentum at point A to the point B as follows:

$\begin{array}{c}m{r}_1{v}_1=m{r}_2{v}_2\\ v_2=\frac{r_1{v}_1}{r_2}\end{array}$

The expression for the kinetic energy at point B $\left(T_1\right)$ as follows:

$T_1=\frac{1}{2}m{v}_1^2$

Here, m is the mass of the satellite.

The expression for the gravitational potential energy at point B $\left(V_1\right)$ as follows:

$V_1=-\frac{GMm}{r_1}$

The expression for the kinetic energy of the orbit at point A $\left(T_2\right)$ as follows:

$T_2=\frac{1}{2}m{v}_2^2$

The expression for the gravitational potential energy at point A $\left(V_2\right)$ as follows:

$V_2=-\frac{GMm}{r_2}$

The expression for the principle of conservation of energy at the point A to point P as follows:

$T_1+V_1=T_2+V_2$

Substitute $\frac{1}{2}m{v}_1^2$ for $T_1$, $-\frac{GMm}{r_1}$ for $V_1$, $\frac{1}{2}m{v}_2^2$ for $T_2$, and $-\frac{GMm}{r_2}$ for $V_2$.

$\begin{array}{l}\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}\\ \frac{v_1^2}{2}-\frac{GM}{r_1}=\frac{v_2^2}{2}-\frac{GM}{r_2}\end{array}$

Substitute $\frac{r_1{v}_1}{r_2}$ for $v_2$.

$\begin{array}{l}\frac{v_1^2}{2}-\frac{GM}{r_1}=\frac{v_2^2}{2}-\frac{GM}{r_2}\\ \frac{v_1^2}{2}-\frac{GM}{r_1}=\frac{{\left(\frac{r_1{v}_1}{r_2}\right)}^2}{2}-\frac{GM}{r_2}\\ \frac{v_1^2}{2}-\frac{GM}{r_1}=\frac{r_1^2{v}_1^2}{2r_2^2}-\frac{GM}{r_2}\end{array}$

$\begin{array}{l}\frac{v_1^2{r}_1-2GM}{2r_1}=\frac{r_1^2{v}_1^2-2GM{r}_2}{2r_2^2}\\ \frac{v_1^2{r}_1-2GM}{r_1}-\frac{r_1^2{v}_1^2-2GM{r}_2}{r_2^2}=0\\ \frac{v_1^2{r}_1}{r_1}-\frac{2GM}{r_1}-\frac{r_1^2{v}_1^2}{r_2^2}+\frac{2GM{r}_2}{r_2^2}=0\end{array}$

Simplify the Equation,

$\begin{array}{l}{v}_1^2-\frac{2GM}{r_1}-{\left(\frac{r_1}{r_2}\right)}^2{v}_1^2+\frac{2GM}{r_2}=0\\ v_1^2\left[1-{\left(\frac{r_1}{r_2}\right)}^2\right]=\frac{2GM}{r_1}-\frac{2GM}{r_2}\\ v_1^2\left(\frac{r_2^2-r_1^2}{r_2^2}\right)=2GM\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\end{array}$

$\begin{array}{l}{v}_1^2\left(\frac{\left(r_2+r_1\right)\left(r_2-r_1\right)}{r_2^2}\right)=2GM\left(\frac{r_2-r_1}{r_2{r}_1}\right)\\ v_1^2\left(\frac{\left(r_2+r_1\right)}{r_2}\right)=2GM\left(\frac{1}{r_1}\right)\\ v_1^2=2GM\left(\frac{1}{r_1}\right)\times \frac{r_2}{\left(r_2+r_1\right)}\\ v_1^2=\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_2}{r_1}\right)\end{array}$

Substitute $\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_2}{r_1}\right)$ for $v_1^2$.

$\begin{array}{c}{v}_2=\frac{r_1\left[\sqrt{\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_2}{r_1}\right)}\right]}{r_2}\\ v_2^2={\left(\frac{r_1}{r_2}\right)}^2\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_2}{r_1}\right)\\ =\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_1}{r_2}\right)\end{array}$

Calculate the kinetic energy in the semi elliptic path AB $\left(T_1\right)$ using the formula:

$T_1=\frac{1}{2}m{v}_1^2$

Substitute $\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_2}{r_1}\right)$ for $v_1^2$.

$\begin{array}{c}{T}_1=\frac{1}{2}m\left(\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_2}{r_1}\right)\right)\\ =\frac{1}{2}m\frac{2GM{r}_2}{r_1\left(r_2+r_1\right)}\end{array}$

Calculate the additional energy $\left(\Delta E_A\right)$ at the point A using the relation:

$\Delta E_A=T_1-T_{circ}$

Substitute $\frac{1}{2}m\frac{2GM{r}_2}{r_1\left(r_2+r_1\right)}$ for $T_1$ and $\frac{GMm}{2r_1}$ for $T_{circ}$.

$\begin{array}{c}\Delta E_A=\frac{1}{2}m\frac{2GM{r}_2}{r_1\left(r_2+r_1\right)}-\frac{GMm}{2r_1}\\ =\frac{2GMm{r}_2-GMm\left(r_2+r_1\right)}{2r_1\left(r_2+r_1\right)}\\ =\frac{GMm\left(2r_2-r_2-r_1\right)}{2r_1\left(r_2+r_1\right)}\\ =\frac{GMm\left(r_2-r_1\right)}{2r_1\left(r_2+r_1\right)}\end{array}$

Divide and Multiply by $r_2$.

$\begin{array}{c}\Delta E_A=\frac{GMm\left(r_2-r_1\right)}{2r_1\left(r_2+r_1\right)}\times \frac{r_2}{r_2}\\ =\frac{r_2}{\left(r_2+r_1\right)}\left(\frac{GMm\left(r_2-r_1\right)}{2r_1{r}_2}\right)\end{array}$

Substitute $\frac{GMm\left(r_2-r_1\right)}{2r_1{r}_2}$ for $\Delta E$.

$\Delta E_A=\frac{r_2}{\left(r_2+r_1\right)}\left(\Delta E\right)$

Calculate the kinetic energy in the semi elliptic path AB $\left(T_2\right)$ using the formula:

$T_2=\frac{1}{2}m{v}_2^2$

Substitute $\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_1}{r_2}\right)$ for $v_2^2$.

$\begin{array}{c}{T}_2=\frac{1}{2}m\left(\frac{2GM}{\left(r_2+r_1\right)}\left(\frac{r_1}{r_2}\right)\right)\\ =\frac{1}{2}m\frac{2GM{r}_1}{r_2\left(r_2+r_1\right)}\end{array}$

Calculate the additional energy $\left(\Delta E_B\right)$ at the point B using the relation:

$\Delta E_B=T_2-T_{circ}$

Substitute $\frac{1}{2}m\frac{2GM{r}_2}{r_1\left(r_2+r_1\right)}$ for $T_2$ and $\frac{GMm}{2r_1}$ for $T_{circ}$.

$\begin{array}{c}\Delta E_B=\frac{1}{2}m\frac{2GM{r}_1}{r_2\left(r_2+r_1\right)}-\frac{GMm}{2r_1}\\ =\frac{2GMm{r}_1-GMm\left(r_2+r_1\right)}{2r_2\left(r_2+r_1\right)}\\ =\frac{GMm\left(2r_1-r_2-r_1\right)}{2r_2\left(r_2+r_1\right)}\\ =\frac{GMm\left(r_1-r_2\right)}{2r_2\left(r_2+r_1\right)}\end{array}$

Divide and Multiply by $r_1$.

$\begin{array}{c}\Delta E_B=\frac{GMm\left(r_1-r_2\right)}{2r_2\left(r_2+r_1\right)}\times \frac{r_1}{r_1}\\ =\frac{r_1}{\left(r_2+r_1\right)}\left(\frac{GMm\left(r_1-r_2\right)}{2r_1{r}_2}\right)\end{array}$

Substitute $\frac{GMm\left(r_2-r_1\right)}{2r_1{r}_2}$ for $\Delta E$.

$\Delta E_B=\frac{r_1}{\left(r_2+r_1\right)}\left(\Delta E\right)$

Therefore, the amount of energy imparted at A $\left(\Delta E_A\right)$ and B $\left(\Delta E_B\right)$ are $\underset{\_}{\Delta E_A=\frac{r_2}{r_1+r_2}\Delta E}$ and $\underset{\_}{\Delta E_B=\frac{r_1}{r_1+r_2}\Delta E}$ respectively and hence it is proved.