Question

Two small spaceships, each with mass$\mathbf{m}\mathbf{=}\mathbf{2000}\mathbf{\text{kg}}$, are in the circular Earth orbit of the figure, at an altitude$\mathbf{\text{h}}$of$\mathbf{400}\mathbf{\text{km}}$.Igor, the commander of one of the ships, arrives at any fixed point in the orbit$\mathbf{90}\mathbf{\text{s}}$ahead of Picard, the commander of the other ship. What are the (a) period${\mathbf{T}}_{\mathbf{0}}$and (b) speed${\mathbf{v}}_{\mathbf{0}}$of the ships? At point P in the figure, Picard fires an instantaneous burst in the forward direction, reducing his ship’s speed by$\mathbf{1.00}\mathit{\%}\mathbf{.}$after this burst; he follows the elliptical orbit shown dashed in the figure. What are the(c) kinetic energy and (d) potential energy of his ship immediately after the burst? In Picard’s new elliptical orbit, what are (e) the total energy$\mathbf{E}$,(f) the semi major axisrole="math" localid="1661171269628" $\mathbf{a}$, and(g) the orbital period$\mathbf{T}$?(h) How much earlier than Igor will Picard return toP?

Short answer

a)Theperiod of the shipsis$92.3\min$

b)Thespeed of the ships is$7.68\times {10}^3\mathrm{m}/{\mathrm{s}}^2$

c)The new kinetic energy is$5.78\times {10}^{10}\mathrm{J}$

d)The new potential energy is$-1.18\times {10}^{11}\mathrm{J}$

e)In the new elliptical orbit, the total energy is $-6.02\times {10}^{10}\mathrm{J}$.

f) The semi-major axis is $6.63\times {10}^6\mathrm{m}$.

g) The time period when ris replaced with ais$89.5\min$

h)Picard will arrive back at point P ahead of Igor by$80\mathrm{s}$

Step-by-step solution

Listing the given quantities

Altitude of$\mathrm{h}=400\mathrm{km}$

Mass of the spaceship$\mathrm{m}=2000\mathrm{kg}$

Understanding the concept of orbital motion

Here, the concept of the period, velocity is used

Formula:$\mathrm{T}0=\sqrt{\frac{4{\mathrm{\pi }}^2{\mathrm{r}}^3}{\mathrm{GM}}}$

(a) Calculation of the period of the ships

The orbital radius

$\begin{array}{c}\mathrm{r}=\mathrm{RE}+\mathrm{h}\\ =6370\mathrm{km}+400\mathrm{km}\\ =6770\mathrm{km}\\ =6.77\times {10}^6\mathrm{m}\end{array}$

Using Kepler's law given in Eq.13-34, we find the period of the ships

$\begin{array}{c}\mathrm{T}0=\sqrt{\frac{4{\mathrm{\pi }}^2{\mathrm{r}}^3}{\mathrm{GM}}}\\ =\sqrt{\frac{4{\mathrm{\pi }}^2{(6.77\times {10}^6\mathrm{m})}^3}{(6.67\times {10}^{-11}{\mathrm{m}}^3/{\mathrm{s}}^2.\mathrm{kg})(5.98\times {10}^{-24}\mathrm{kg})}}\\ =(5.54\times {10}^3\mathrm{s})\\ \approx 92.3\min \end{array}$

The period of the ships is$92.3\min$

(b) Calculation of the speed of the ships

The speed of the ships is

$\begin{array}{c}\mathrm{v}0=\frac{2\mathrm{\pi r}}{\mathrm{T}0}\\ =\frac{2\mathrm{\pi }(6.77\times {10}^6\mathrm{m})}{5.54\times {10}^3\mathrm{s}}\\ =7.68\times {10}^3\mathrm{m}/{\mathrm{s}}^2\end{array}$

The speed of the ships is$7.68\times {10}^3\mathrm{m}/{\mathrm{s}}^2$

(c) Calculation of the new kinetic energy

The new kinetic energy is

$\begin{array}{l}\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2\\ =\frac{1}{2}\mathrm{m}{\left(0.99{\mathrm{V}}_0\right)}^2\\ =\frac{1}{2}\left(2000\mathrm{kg}\right){\left(0.99\right)}^2{\left(7.68\times {10}^3\mathrm{m}/\mathrm{s}\right)}^2\\ =5.78\times {10}^{10}\mathrm{J}\end{array}$

The new kinetic energy is$5.78\times {10}^{10}\mathrm{J}$

(d) Calculation of the potential energy 

Immediately after the burst, the potential energy is the same as it was before the burst. Therefore,

$\begin{array}{c}\mathrm{U}=\frac{\mathrm{GMm}}{\mathrm{r}}\\ =\frac{(6.77\times {10}^{-11}{\mathrm{m}}^3/{\mathrm{s}}^2.\mathrm{kg})(5.98\times {10}^{24}\mathrm{kg})\left(2000\mathrm{kg}\right)}{6.77\times {10}^6\mathrm{m}}\\ =-1.18\times {10}^{11}\mathrm{J}\end{array}$

The new potential energy is$-1.18\times {10}^{11}\mathrm{J}$

(e) Calculation of the total energy

In the new elliptical orbit, the total energy is

$\begin{array}{c}\mathrm{E}=\mathrm{K}+\mathrm{U}\\ =\frac{(5.78\times {10}^{10}\mathrm{J})+(-1.18\times {10}^{11}\mathrm{J})}{6.77\times {10}^6\mathrm{m}}\\ =-6.02\times {10}^{10}\mathrm{J}\end{array}$

In the new elliptical orbit, the total energy is $-6.02\times {10}^{10}\mathrm{J}$.

(f) Calculation of the semi-major axis 

For elliptical orbit, the total energy can be written as (see Eq. 13-42) $\mathrm{E}=\frac{-\mathrm{GMm}}{2\mathrm{a}}$where a is the semi-major axis. Thus,

$\begin{array}{c}\mathrm{a}=\frac{-\mathrm{GMm}}{2\mathrm{E}}\\ =\frac{(6.77\times {10}^{-11}{\mathrm{m}}^3/{\mathrm{s}}^2.\mathrm{kg})(5.98\times {10}^{24}\mathrm{kg})\left(2000\mathrm{kg}\right)}{2(-6.02\times {10}^{10}\mathrm{J})}\\ =6.63\times {10}^6\mathrm{m}\end{array}$

The semi-major axis is$6.63\times {10}^6\mathrm{m}$.

(g) Calculation of the period

The period, we use Eq. 13-34 but replace r with a. The result is

$\begin{array}{c}\mathrm{T}=\sqrt{\frac{4{\mathrm{\pi }}^2{\mathrm{a}}^3}{\mathrm{GM}}}\\ =\sqrt{\frac{4{\mathrm{\pi }}^2{(6.63\times {10}^6\mathrm{m})}^3}{(6.67\times {10}^{-11}{\mathrm{m}}^3/{\mathrm{s}}^2.\mathrm{kg})(5.98\times {10}^{24}\mathrm{kg})}}\\ =(5.37\times {10}^3\mathrm{s})\\ \approx 89.5\min \end{array}$

The time period when ris replaced with ais$89.5\min$

(h) Explanation

The orbital period T for Picard's elliptical orbit is shorter than Igor's by

$\begin{array}{c}\mathrm{\Delta T}={\mathrm{T}}_0-\mathrm{T}\\ =5540\mathrm{s}-5370\mathrm{s}\\ =170\mathrm{s}\end{array}$

$170\mathrm{s}-90\mathrm{s}=80\mathrm{s}$

Thus, Picard will arrive back at point P ahead of Igor by$80\mathrm{s}$