91Ó°ÊÓ

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

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

Here, the concept of the period, velocity is used

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

The orbital radius

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

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

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

The period of the ships is$92.3\text{ }\min$

The speed of the ships is

$\begin{array}{c}\mathrm{v}0=\frac{2\mathrm{Ï€°ù}}{\mathrm{T}0}\\ =\frac{2\mathrm{Ï€}(6.77×{10}^6\mathrm{m})}{5.54×{10}^3\mathrm{s}}\\ =7.68×{10}^3\mathrm{m}/{\mathrm{s}}^2\end{array}$

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

The new kinetic energy is

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

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

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

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

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

In the new elliptical orbit, the total energy is

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

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

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,

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

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

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

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

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

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

$\begin{array}{c}\mathrm{Δ°Õ}={\mathrm{T}}_0-\mathrm{T}\\ =5540\text{ }\mathrm{s}-5370\text{ }\mathrm{s}\\ =170\text{ }\mathrm{s}\end{array}$

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

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