Chapter 5, Problem 85GP

(a)

To determine

The time to overtake the satellite.

Answer to Problem 85GP

The time to overtake the satellite is $2000\text{s}$ .

Explanation of Solution

Given info:

The radius of the circular orbit is, $r_o=400\text{km}=4\times {10}^5\text{m}$ .

The distance between the space shuttle and satellite is, $d=25\text{km}$ .

The reduction in the radius of the orbit is, $r=1\text{km}=1000\text{km}$ .

Formula Used:

The expression to calculate the relative velocity of the space shuttle is,

  $v=\sqrt{GM}\left(\frac{1}{\sqrt{R+r_o}}-\frac{1}{\sqrt{R+r_o-r}}\right)$

Here,

• G is the universal gravitational constant and its value is $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ .
• M is the mass of the earth and its value is $5.98\times {10}^{24}\text{kg}$ .
• R is the radius of the earth and its value is $6.38\times {10}^6\text{km}$ .

The expression to calculate the time to overtake the satellite is,

  $T=\frac{r}{v}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}v=\left[\sqrt{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)}\left(\begin{array}{l}\frac{1}{\sqrt{6.38\times {10}^6\text{km}+400\text{km}}}-\\ \frac{1}{\sqrt{6.38\times {10}^6\text{km}+400\text{km}-1\text{km}}}\end{array}\right)\right]\\ =0.5\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}T=\frac{1000\text{m}}{0.5\text{m}/\text{s}}\\ =2000\text{s}\end{array}$

Conclusion:

Thus, the time to overtake the satellite is $2000\text{s}$ .

(b)

To determine

The reduction in the orbital radius.

Answer to Problem 85GP

The reduction in radius is $100\text{m}$ .

Explanation of Solution

Given info:

The time in the orbit is, $t=7\text{h}=7\text{h}\times \frac{3600\text{s}}{1\text{h}}=25200\text{s}$ .

Formula Used:

The expression to calculate the change in speed is,

  $u=\frac{r}{t}$

The expression to calculate the speed of the satellite is,

  $u_{sat}=\sqrt{\frac{GM}{r_o+R}}$

The expression to calculate the speed of the astronaut is,

  $v_{ast}=u+u_{sat}$

The expression to calculate the radius of the orbit for astronaut is,

  $r_{ast}=\frac{GM}{v_{ast}^2}$

The expression to calculate the reduction in radius is,

  $x=r_0-\left(r_{ast}-R\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}u=\frac{1000\text{m}}{25200\text{}\text{s}}\\ =0.039\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{u}_{sat}=\sqrt{\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)}{4\times {10}^5\text{m}+6.38\times {10}^6}}\\ =7670.1\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{v}_{ast}=7670.1\text{m}/\text{s}+0.039\text{m}/\text{s}\\ =7670.139\text{m}/\text{s}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{r}_{sat}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(7670.139\text{m}/\text{s}\right)}^2}\\ =6.7799\times {10}^6\text{m}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}x=4\times {10}^5\text{m}-\left(6.7799\times {10}^6\text{m}-6.38\times {10}^6\right)\\ =100\text{m}\end{array}$

Conclusion:

Thus, the reduction in radius is $100\text{m}$ .