Chapter 5, Problem 86GP

(a)

To determine

To find : The mean distance from Sun.

Answer to Problem 86GP

  $30.9\times {10}^9$ km

Explanation of Solution

Given:

  $T=3000$ years

Formula Used :

Time period is given as,

  $T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$

Calculation :

Substituting the given value

  $9.46\times {10}^{10}=\frac{2\pi r^{3/2}}{\sqrt{6.67\times {10}^{-11}\times 1.98\times {10}^{30}}}$

  $\Rightarrow r=30.9\times {10}^9$ km

Conclusion :

Therefore the mean distance is $30.9\times {10}^9$ km.

(b)

To determine

To find : The farthest distance.

Answer to Problem 86GP

  $6.65\times {10}^9$ km

Explanation of Solution

Given:

Minimum distance $=0.15\times {10}^9$ km

Formula Used :

  $\text{Mean distance =}\frac{\text{Minimum distance + Maximum distance}}{2}$

Calculation :

Substituting the given values

  $30.9\times {10}^9=\frac{0.15\times {10}^9+\text{Maximum distance}}{2}$

  $\Rightarrow \text{ Maximum distance = 61}{\text{.65×10}}^9$ km

Conclusion :

Therefore the maximum distance is $\text{61}{\text{.65×10}}^9$ km.

(c)

To determine

To find : The ratio of speed at closet point to the farthest point.

Answer to Problem 86GP

4.11

Explanation of Solution

Given:

  $r_1=0.15\times {10}^9$ km

  $r_2=61.65\times {10}^9$ km

Formula Used :

The speed ans distance are related as,

  $\frac{v_1}{v_2}=\frac{r_2}{r_1}$

Calculation :

Substituting the values

  $\frac{v_1}{v_2}=\frac{61.65\times {10}^9}{0.15\times {10}^9}=4.11$

Conclusion :

Therefore the ratio of speed is 4.11.