Question

Question: In a shuttle craft of mass m = 3.00 kg , Captain Janeway orbits a planet of mass $\mathit{M}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{25}}\mathit{k}\mathit{g}$ , in a circular orbit of radius$\mathit{r}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathit{m}$ .What are (a) the period of the orbit and (b) the speed of the shuttle craft? Janeway briefly fires a forward pointing thruster, reducing her speeds by 2.00%. Just then, what are (c) the speed, (d) the kinetic energy, (e) the gravitational potential energy, and (f) the mechanical energy of the shuttle craft? (g) What is the semi major axis of the elliptical orbit now taken by the craft? (h)What is the difference between the period of the original circular orbit and that of the new elliptical orbit? (i) Which orbit has the smaller period?

Short answer

Answer:

1. The period of orbit is$T=2.15\times {10}^{4}s$
2. The speed of the shuttle craft is$v=1.23\times {10}^{4}m/s$
3. The speed of the shuttle craft (after firing) is$v\text{'}=1.20\times {10}^{4}m/s$
4. The kinetic energy (after firing) is$K=2.17\times {10}^{11}J$
5. The gravitational potential energy (after firing) is$U=-4.53\times {10}^{11}J$
6. The mechanical energy of the shuttle craft is$E=-2.36\times {10}^{11}J$
7. The semi major axis of the elliptical orbit is$a=4.04\times {10}^{7}m$
8. The difference between the period of the original circular orbit and new elliptical orbit is$T-T\text{'}=1.22\times {10}^{3}s$
9. The elliptical orbit has a smaller period.

Step-by-step solution

Identification of given data

The mass of the craft is $m=3000kg$

The mass of the planet is $M=9.50\times {10}^{25}kg$

The radius of the circular orbit is $r=4.20\times {10}^{7}m$

Significance of Kepler’s law of periods

The square of a planet's period of revolution around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, states Kepler's law of periods.

We can use the concept of Kepler’s law of period and expressions of the period, speed of shuttle craft, gravitational potential energy, mechanical, and semi major axis of elliptical orbit.

Formula

$T^2=\frac{4{\pi }^2}{GM}{r}^3$

$$\begin{gathered} v=\frac{2\pi r}{T} \\ K=\frac{1}{2}m{v}^2 \\ U=-\frac{GMm}{r} \\ E=-\frac{GMm}{2a} \end{gathered}$$

Where,

K is the kinetic energy, U is the potential energy, E is the mechanical energy of object, is the time period of object, T is the gravitational G constant $6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$, M is the mass of planet , v is the speed of object, m is the mass of object, r is the orbital radius of object and a is semi major axis of the elliptical orbit

(a) Determining the period of orbit

According to Kepler’s law of period,

$$\begin{gathered} T^2=\frac{4{\pi }^2}{GM}{r}^3 \\ T=\sqrt{\frac{4{\pi }^2\times {\left(4.20\times {10}^{7}m\right)}^3}{{\left(6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·\text{s}\right)}^{\text{2}}\left(9.50\times {10}^{25}kg\right)}} \\ =2.15\times {10}^{4}s \end{gathered}$$

The period of orbit is $T=2.15\times {10}^{4}s$

(b) Determining the speed of shuttle craft

According to the expression of orbital speed,

$$\begin{gathered} v=\frac{2\pi r}{T} \\ =\frac{2\times 3.142\times 4.20\times {10}^7\text{m}}{2.15\times {10}^4\text{s}} \\ =1.23\times {10}^4\text{m/s} \end{gathered}$$

The speed of the shuttle craft is $v=1.23\times {10}^4\text{m/s}$

(c) Determining the speed after firing 

After firing the thruster, the speed reduces by two percent. Hence,

$$\begin{gathered} v\text{'}=0.98 v \\ =0.98\times 1.23\times {10}^4\text{m/s} \\ =1.20\times {10}^4\text{m/s} \end{gathered}$$

The speed of the shuttle craft (after firing) is $v\text{'}=1.20\times {10}^4\text{m/s}$

(d) Determining the kinetic energy

The expression of kinetic energy is

$$\begin{gathered} K=\frac{1}{2}m{\left(v\text{'}\right)}^2 \\ =\frac{1}{2}\times 3000\text{kg}\times {\left(1.203\times {10}^4\text{m/s}\right)}^2 \\ =2.17\times {10}^{11}\text{J} \end{gathered}$$

The kinetic energy (after firing) is $K=2.17\times {10}^{11}J$

(e) Determining the gravitational potential energy

$$\begin{gathered} U=-\frac{GMm}{r} \\ =-\frac{\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(9.50\times {10}^{25}\text{kg}\right)\times 3000 \text{kg}}{4.20\times {10}^7\text{m}} \\ =-4.53\times {10}^{11}\text{J} \end{gathered}$$

The gravitational potential energy (after firing) is $U=-4.53\times {10}^{11}J$

(f) Determining the mechanical energy of shuttle craft 

The mechanical energy of the shuttle craft:

$$\begin{gathered} E=K+U \\ =2.17\times {10}^{11}\text{J}+\left(-4.53\times {10}^{11}\text{J}\right) \\ =-2.36\times {10}^{11}\text{J} \end{gathered}$$

The mechanical energy of the shuttle craft is $E=-2.36\times {10}^{11}J$

(g) Determining the semi major axis of the elliptical orbit

$$\begin{gathered} a=-\frac{GMm}{2E} \\ =-\frac{\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(9.50\times {10}^{25}\text{kg}\right)\times 3000\text{kg}}{2\times \left(-2.36\times {10}^{11}\text{J}\right)} \\ =4.04\times {10}^7\text{m} \end{gathered}$$

The semi major axis of the elliptical orbit is $a=4.04\times {10}^{7}m$

(h) Determining the difference between the period of the original circular orbit and new elliptical orbit:

By using Kepler’s law of period, we can find the new period for the elliptical orbit as

$$\begin{gathered} T^{\text{'}2}=\frac{4{\pi }^2}{GM}{a}^3 \\ T\text{'}=\sqrt{\frac{4{\pi }^2}{GM}{a}^3} \\ =\sqrt{\frac{4\times {\left(3.142\right)}^2\times {\left(4.04\times {10}^7\text{m}\right)}^3}{\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(9.50\times {10}^{25}\text{kg}\right)}} \\ =2.03\times {10}^4\text{s} \\ T-T\text{'}=2.15\times {10}^4\text{s}-2.03\times {10}^4\text{s} \\ =1.22\times {10}^3\text{s} \end{gathered}$$

The difference between the period of the original circular orbit and new elliptical orbit is

$T-T\text{'}=1.22\times {10}^3\text{s}$

(i) Explanation of comparison of the period of circular orbit with that of elliptical orbit: 

By comparing the period of circular orbit with the period of elliptical orbit, we find that the period of elliptical orbit is smaller.