91Ó°ÊÓ

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

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

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

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

$T^2=\frac{4{\mathrm{Ï€}}^2}{GM}{r}^3$

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

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 $6.67×{10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$, 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

According to Kepler’s law of period,

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

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

According to the expression of orbital speed,

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

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

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

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

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

The expression of kinetic energy is

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

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

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

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

The mechanical energy of the shuttle craft:

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

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

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

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

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

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

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

$T-T\text{'}=1.22×{10}^3\text{s}$

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