Question

If earth suddenly contracts to half its present radius keeping the mass constant, what would be the length of the day?

Answer 1

Step 1. Given data:

From the given, reduction in Earth’s size indicates the reduction in its radius, since, the radius is reduced to half.

Step 2. Formula used:

We have to find the moment of inertia of earth before and after contraction keeping in mind that the mass remains constant.

The formula of the moment of inertia of a solid sphere is,

$\mathrm{I}=\frac{2}{5}{\mathrm{MR}}^2$

Here $\mathrm{M}$ is the mass and $\mathrm{R}$ is the radius of the sphere.

The angular velocity can be expressed as,

$\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}$

Now the law of conservation of angular momentum is,

${\mathrm{I}}_1{\mathrm{\omega }}_1={\mathrm{I}}_2{\mathrm{\omega }}_2$

From the given, the present radius of the earth ${\mathrm{R}}_1$ is reduced by half, so after contraction the radius is,

${\mathrm{R}}_2=\frac{{\mathrm{R}}_1}{2}$ ------------ $\left(1\right)$

But at the time of contraction, the mass is constant, which is shown in the figure above

We get,

${\mathrm{M}}_1={\mathrm{M}}_2=\mathrm{M}$ ------------ $\left(2\right)$

Step 3. Find the angular velocity before and after contraction

From the above contraction, the angular velocity of the earth's rotation changes.

So, the angular velocity before contraction is,

${\mathrm{\omega }}_1=\frac{2\mathrm{\pi }}{{\mathrm{T}}_1}$----------- $\left(3\right)$

The angular velocity after contraction is,

${\mathrm{\omega }}_2=\frac{2\mathrm{\pi }}{{\mathrm{T}}_2}$---------- $\left(4\right)$

Step 4. Find the moment of inertia

There is another quantity that is undergoing change due to the contraction of earths- moment of inertia. So, the moment of inertial before contraction is,

${\mathrm{I}}_1=\frac{2}{5}{{\mathrm{MR}}_1}^2$ ----------- $\left(5\right)$

Moment of inertia after contraction is,

${\mathrm{I}}_2=\frac{2}{5}{{\mathrm{MR}}_2}^2$

$=\frac{2}{5}M{\left(\frac{{\mathrm{R}}_1}{2}\right)}^2$

$=\frac{{\mathrm{I}}_1}{4}$ -------- $\left(6\right)$

Now,

According to the law of conservation of angular momentum is,

$\frac{\mathrm{dL}}{\mathrm{dt}}=0$

Here, $\mathrm{L}=\mathrm{I\omega }$ is constant.

So,

${\mathrm{I}}_1{\mathrm{\omega }}_1={\mathrm{I}}_2{\mathrm{\omega }}_2$ --------- $\left(7\right)$

Step 5. Find the length of the day

Substitute equations $\left(3\right)$, $\left(4\right)$, and $\left(6\right)$ in $\left(7\right)$, we get,

${\mathrm{I}}_1\left(\frac{2\mathrm{\pi }}{{\mathrm{T}}_1}\right)=\left(\frac{{\mathrm{I}}_1}{4}\right)\left(\frac{2\mathrm{\pi }}{{\mathrm{T}}_2}\right)$

${\mathrm{T}}_2=\frac{{\mathrm{T}}_1}{4}$ ----------- $\left(8\right)$

We know that our planet Earth rotates once every $24$ hours at the moment.

So,

${\mathrm{T}}_1=24\mathrm{hrs}$ --------- $\left(9\right)$

Substitute the equation $\left(9\right)$ in $\left(8\right)$, we get,

${\mathrm{T}}_2=\frac{24}{4}\mathrm{Hrs}$

${\mathrm{T}}_2=6\mathrm{Hrs}$

Hence, for an earth that is contracted to half its present size with mass constant, the duration of a day will be just $6\mathrm{Hrs}$.