Chapter 5, Problem 84GP

(a)

To determine

To find: The time period of rotating Asteroid.

Answer to Problem 84GP

  $2\times {10}^4\text{s}$ .

Explanation of Solution

Given:

  $h=15$ km

  $v=40\text{km}\times \text{6km}\times \text{6km}$ km

  $T=2.3\times {10}^3$ kg/m³

Formula Used:

  $T=2\pi h\sqrt{\frac{h}{GM}}$

Calculation:

The speed of the asteroid orbiting a the Eros with radius $r$ is given by $v=\sqrt{\frac{GM}{r}}$

And the time period of the asteroid is given by,

  $T=\frac{2\pi r}{\sqrt{\frac{GM}{r}}}$

Where the Gravitational constant $G=6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$

  $\begin{array}{l}=\frac{2\pi \left(15\times {10}^3\right)}{\sqrt{\frac{\left(6.67\times {10}^{-11}\right)\left(3.312\times {10}^{15}\right)}{\left(15\times {10}^3\right)}}}\\ =2\times {10}^4\text{s}\end{array}$

Conclusion:

Therefore the time period will be $2\times {10}^4\text{s}$ .

(b)

To determine

To find: The radius of sphere.

Answer to Problem 84GP

  $7$ km

Explanation of Solution

Given Data:

  $v=40\text{ km × 6 km × 6 km}$

  $T=2.3\times {10}^3{\text{ kg/m}}^3$

Formula Used:

  $v=\frac{4}{3}\pi r^3$

Calculation:

If the Eros were a sphere with radius $R$

Then the volume of the sphere is given by, $V=\frac{4}{3}\pi R^3$

The mass of the Eros, $M=\rho V$

And

  $\begin{array}{l}V=\frac{M}{\rho }\\ \frac{4}{3}\pi R^3=\frac{M}{\rho }\\ R={\left(\frac{3M}{4\pi \rho }\right)}^{\frac{1}{3}}\\ ={\left(\frac{3\left(3.312\times {10}^{15}\right)}{4\pi \left(2.3\times {10}^3\right)}\right)}^{\frac{1}{3}}\\ =7\times {10}^3\text{m}\end{array}$

Conclusion:

Therefore radius is $7$ km.

(c)

To determine

To find: The acceleration due to gravity.

Answer to Problem 84GP

  $4.5\times {10}^{-3}$ m/s²

Explanation of Solution

Given Data:

  $M=3.31\times {10}^{15}$ kg

  $r=7$ km

Formula Used:

  $g=\frac{GM}{r^2}$

Calculation:

Substituting the values

  $g=\frac{6.67\times {10}^{-11}\times 3.3\times {10}^{15}}{49\times {10}^6}$

  $=4.5\times {10}^{-3}$ m/s²

Conclusion:

Therefore the acceleration due to gravity will be $4.5\times {10}^{-3}$ m/s².