Chapter 7, Problem 96A

To determine

The speed of a planet of Earth’s mass and size to spin so that an object at the equator would be weightless.

To find:The period in minutes.

Answer to Problem 96A

An object at the equator feels weightless when centripetal acceleration of the Earth equals to the acceleration due to gravity of Earth.

  $\text{T= 84}\text{.5 min}$

Explanation of Solution

Introduction:

The planet Earth accelerates an object on its surface in two directions. Acceleration due to gravity of Earth pulls an object towards its center and centripetal acceleration pushes an object away from the axis of rotation of Earth. Centripetal acceleration is the acceleration which acts on the object that undergoes circular motion. Centripetal acceleration of Earth also has an effect on the weight of an object on the surface of planet Earth in such a way that an object at the equator feel slightly less weight compared to that at the pole of the Earth.

When centripetal acceleration of the Earth and acceleration due to gravity of Earth become equal, then the surface of the Earth would not have to supply any force on an object on it. At this situation, an object at the equator feels weightless.

To find the period:

If an object moving in a circular orbit of radius $r$ with a constant speed $v$ , then its centripetal acceleration is given by,

  $a_c=\frac{v^2}{r}$$\left(1\right)$

Earth’s acceleration due to gravity is given by,

  $g=\frac{GM}{r^2}$$\left(2\right)$

Where, $M$ is mass of the Erath, $M=5.97\times {10}^{24}\text{ kg}$ $r$ is radius of Erath, $\text{r = 6}\text{.38}\times {\text{10}}^6\text{ m}$ $\text{G}$ is gravitational constant, $\text{G = 6}\text{.67}\times {\text{10}}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^{\text{2}}$

When centripetal acceleration of the Earth equals to the acceleration due to gravity of Earth,

  $\frac{v^2}{r}=\frac{GM}{r^2}$

  $\Rightarrow v^2=\frac{GM}{r}\Rightarrow v=\sqrt{\frac{GM}{r}}$$\left(3\right)$

But, $v=\frac{2\pi r}{T}$

  $\Rightarrow T=\frac{2\pi r}{v}$$\left(4\right)$

Where, $T$ is the period of revolution

Substituting for $v$ from equation $\left(3\right)$ ,

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

  $\Rightarrow T=2\pi \sqrt{\frac{r^3}{GM}}$$\left(5\right)$

Substituting the numerical values in equation $\left(5\right)$ ,

  $T=2\pi \sqrt{\frac{{\left(\text{6}\text{.38}\times {\text{10}}^6\text{ m}\right)}^3}{\left(\text{6}\text{.67}\times {\text{10}}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.97\times {10}^{24}\text{ kg}\right)}}$

  $=2\pi \sqrt{\frac{259.69\times {10}^{18}}{39.81\times {10}^{13}}}$

  $=2\pi \sqrt{65.23\times {10}^4}$

  $=2\pi \times 8.07\times {10}^2$

  $=5.07\times {10}^3\text{ s }$

  $\begin{array}{l}\text{=}\frac{5.07\times {10}^3}{60}\text{ min}\\ \text{ = 84}\text{.5 min}\end{array}$

Conclusion:

An object at the equator feels weightless when centripetal acceleration of the Earth equals to the acceleration due to gravity of Earth.Period of revolution of Earth is $\text{84}\text{.5 min}$ .