Question

A rocket with mass $\mathbf{5}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\text{\hspace{0.17em}kg}}$is in a circular orbit of radius$\mathbf{7}\mathbf{.20}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{\hspace{0.17em}m}}$around the earth. The rocket’s engines fire for a period of time to increase that radius to$\mathbf{8}\mathbf{.80}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{\hspace{0.17em}m}}$, with the orbit again circular. (a) What is the change in the rocket’s kinetic energy? Does the kinetic energy increase or decrease? (b) What is the change in the rocket’s gravitational potential energy? Does the potential energy increase or decrease? (c) How much work is done by the rocket engines in changing the orbital radius?

Short answer

1. The change in the kinetic energy decreases and its value is, $2.52\times {10}^{10}\text{\hspace{0.17em}J}$.
2. The change in potential energy increases and its value is, $5.04\times {10}^{10}\text{\hspace{0.17em}J}$.
3. The work required to change its orbit is,$\text{2.51}\times {10}^{10}\text{\hspace{0.17em}J}$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The mass of the rocket is,$m=5.00\times {10}^3\text{\hspace{0.17em}kg}$.
• The radius of the rocket is,$r_1=7.20\times {10}^6\text{\hspace{0.17em}m}$.
• After the fire the radius of the rocket is,$r_2=8.80\times {10}^6\text{\hspace{0.17em}m}$.

Significance of conservation of energy

The total amount of energy of an isolated system remains stable, even though internal changes occur when energy disappears in one form and appears in another.

Determination of the change in kinetic

a)

The rocket move in a circular orbit, then the speed of the rocket is expressed as,

$v=\sqrt{\frac{G{M}_E}{r}}$

Here$v$is the speed of the rocket,$G$is the gravitational constant, $M_E$is the mass of the earth, and$r$is the radius of the rocket.

The relation of kinetic energy is expressed as,

$K=\frac{1}{2}m{v}^2$ ...(i)

Here K is the kinetic energy and m is the mass of the rocket.

Substitute the value of$v$in equation (i)

$\begin{array}{c}K=\frac{1}{2}m{\left(\sqrt{\frac{G{M}_E}{r}}\right)}^2\\ =\frac{Gm{M}_E}{2r}\end{array}$

The change in kinetic energy is expressed as,

$\begin{array}{c}{K}_2-K_1=\frac{Gm{M}_E}{2r_2}-\frac{Gm{M}_E}{2r_1}\\ =\frac{Gm{M}_E}{2}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\\ =Gm{M}_E\left(\frac{r_1-r_2}{2r_1{r}_2}\right)\end{array}$

Here $K_2$is the kinetic energy after an engine fire, $K_1$is the kinetic energy before engine fire, $r_1$is the radius of the rocket before engine fire, $r_2$is the radius of the rocket after engine fire $G$is the gravitational constant, and $M_E$is the mass of earth’s.

Substitute $6.673\times {10}^{-11}{\text{\hspace{0.17em}Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$for $G$, $5000kg$for $m$, $5.97\times {10}^{24}\text{\hspace{0.17em}kg}$for $M_E$, $7.20\times {10}^6\text{\hspace{0.17em}m}$for $r_1$,and $8.80\times {10}^6\text{\hspace{0.17em}m}$ for $r_2$in the above equation.

$\begin{array}{c}{K}_2-K_1=6.673\times {10}^{-11}{\text{\hspace{0.17em}Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5000\text{\hspace{0.17em}kg}\times 5.97\times {10}^{24}\text{\hspace{0.17em}kg}\times \left(\frac{7.20\times {10}^6\text{\hspace{0.17em}m}-8.80\times {10}^6\text{\hspace{0.17em}m}}{2\times 7.20\times {10}^6\text{\hspace{0.17em}m}\times 8.80\times {10}^6\text{\hspace{0.17em}m}}\right)\\ K_2-K=-2.52\times {10}^{10}\text{\hspace{0.17em}J}\end{array}$

Hence the change in the kinetic energy decreases and its value is, $2.52\times {10}^{10}\text{\hspace{0.17em}J}$.

Determination of the change in potential energy

b)

The relation of potential energy is expressed as,

$U=-G\frac{M_{E}m}{r}$

Here$U$is the potential energy,$G$is the gravitational constant,$M_E$is the mass of the earth, and$m$is the mass of the rocket.

The change in gravitational potential energy is expressed as,

$\begin{array}{c}{U}_2-U_1=-\frac{Gm{M}_E}{r_2}+\frac{Gm{M}_E}{r_1}\\ =Gm{M}_E\left(-\frac{1}{r_2}+\frac{1}{r_1}\right)\\ =Gm{M}_E\left(\frac{r_2-r_1}{r_1{r}_2}\right)\end{array}$

Here$U_2$ is the gravitational potential energy after an engine fire, $U_1$is the gravitational potential energy before engine fire.

Substitute $6.673\times {10}^{-11}{\text{\hspace{0.17em}Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$for $G$, $5000Kg$for $m$, $5.97\times {10}^{24}\text{\hspace{0.17em}kg}$for $M_E$, $7.20\times {10}^6\text{\hspace{0.17em}m}$for $r_1$, and $8.80\times {10}^6\text{\hspace{0.17em}m}$ for$r_2$ in the above equation.

$\begin{array}{c}{U}_2-U_1=6.673\times {10}^{-11}{\text{\hspace{0.17em}Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5000\text{\hspace{0.17em}kg}\times 5.97\times {10}^{24}\text{\hspace{0.17em}kg}\times \left(\frac{8.80\times {10}^6\text{\hspace{0.17em}m}-7.20\times {10}^6\text{\hspace{0.17em}m}}{7.20\times {10}^6\text{\hspace{0.17em}m}\times 8.80\times {10}^6\text{\hspace{0.17em}m}}\right)\\ U_2-U_1=5.04\times {10}^{10}\text{\hspace{0.17em}J}\end{array}$

Hence the change in potential energy increases and its value is,$5.04\times {10}^{10}\text{\hspace{0.17em}J}$ .

Determination of the work required to change its orbit  

c)

The conservation of the energy is expressed as,

$\begin{array}{c}{K}_1+U_1+W_0=K_2+U_2\\ W_0=(K_2-K_1)+(U_2-U_1)\end{array}$

Here$W_o$ is the work required.

Substitute the value of$(K_2-K_1)$ and $(U_2-U_1)$in the above equation take from the results part (a) and part (b).

$\begin{array}{c}{W}_0=(K_2-K_1)+(U_2-U_1)\\ =-2.52\times {10}^{10}\text{\hspace{0.17em}J}+5.04\times {10}^{10}\text{\hspace{0.17em}J}\\ \text{=2.51}\times {10}^{10}\text{\hspace{0.17em}J}\end{array}$

Hence the work required to change its orbit is, $\text{2.51}\times {10}^{10}\text{\hspace{0.17em}J}$.