Q23E - University Physics with Modern Physics

Question

The earth has a radius of 6380 km and turns around once on its axis in 24 h. (a) What is the radial acceleration of an object at the earth’s equator? Give your answer in m/s² and as a fraction of g. (b) If a_rad at the equator is greater than g, objects will fly off the earth’s surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth’s rotation have to be for this to occur?

Step-by-Step Solution

Answer

1. The radial acceleration of the object is 0.0337 m/s².
2. The time period of revolution of earth at equator for radial acceleration greater than g is  84.5 min.

Step 1: Identification of given data:

The given data can be listed below.

• The radius of earth is, R = 6380 km
• The time for one revolution is, $T=24h\left(\frac{60min}{1h}\right)\left(\frac{60 s}{1min}\right)=8.64\times {10}^{4}s$

Step 2: Concept/Significance of radial acceleration:

The term "radial component of acceleration" refers to a resultant acceleration component that is parallel to the instantaneous velocity.

Step 3: (a) Determination of the radial acceleration of an object at the earth’s equator:

The velocity of the object is given by,

$v=\frac{∆x}{∆t}$

Here, $∆x$ is the circumference of earth whose value is $2\mathrm{\pi R}$, and $∆t$ is the time of revolution.

Substitute all the values in the above,

$v=\frac{2\mathrm{\pi R}}{T}$

The radial acceleration of the object is given by,

$\begin{array}{rcl}{a}_r& =& \frac{v^2}{R}\\ a_r& =& \frac{{\left({\displaystyle \frac{2\mathrm{\pi R}}{T}}\right)}^2}{R}\\ & =& \frac{4{\mathrm{\pi }}^2\mathrm{R}}{T^2}\end{array}$

Here, v is the velocity of the object and R is the radius of earth.

Substitute all the values in the above expression.

$\begin{array}{rcl}{a}_r& =& \frac{4{\left(3.14\right)}^2\left(6380km\right)}{{\left(8.64\times {10}^{4}s\right)}^2}\left(\frac{{10}^{3}m}{1km}\right)\\ & =& \frac{4\times 6.28\times 6.38\times {10}^{6}m}{{\left(8.64\times {10}^{4}s\right)}^2}\\ & =& 0.0337m/s^2\end{array}$

Thus, the radial acceleration of the object is 0.0337 m/s².

Step 4: (b) Determination of the period of the earth’s rotation for a rad equal to g.

The time period of earth’s revolution is given by,

$\begin{array}{rcl}{T}^2& =& \frac{4{\mathrm{\pi }}^2{\mathrm{R}}^2}{a_r}\\ T& =& \sqrt{\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^2}{a_r}}\\ & =& \frac{2\mathrm{\pi R}}{\sqrt{a_r}}\end{array}$

Here, R is the radius of earth, and a_r is the radial acceleration of the object which is equal to g.

Substitute all the values in the above,

$\begin{array}{rcl}T& =& \frac{2\mathrm{\pi R}}{\sqrt{g}}\\ & =& \frac{2\left(3.14\right)\left(6.38\times {10}^{6}m\right)}{\sqrt{\left(9.8m/s^2\right)}}\\ & =& 5070s\left(\frac{1min}{60 s}\right)\\ & =& 84.5min\end{array}$

Thus, the time period of revolution of earth at equator for radial acceleration greater than   is 84.5min.