Question

Calculate the magnitude of the angular momentum of the Mars in a circular orbit around the sun. The Mars has mass 6.42×1023 kg , radius 3.40×106 m , and orbit radius 2.28×1011 m . The planet completes one rotation on its axis in 24.5 hours and one orbit in 687 days.

Answer 1

Answer:

The magnitude of the angular momentum of the Mars in a circular orbit around the sun is $3.53\times10^{39}\ kg-m^2/s$

Explanation:

Given that,

Mass of mars $M=6.42\times10^{23}\ kg$

Radius $r'= 3.40\times10^{6}$

Orbit radius $r=2.28\times10^{11}\ m$

Time period = 687 days

We need to calculate the magnitude of the angular momentum of the mars

Using formula of angular momentum

$L = I\omega$

Here, $I = mr^2$

$L=mr^2\omega$

$L=mr^2\times\dfrac{2\pi}{T}$

Where,

L = angular momentum

I = moment of inertia

r = radius

$\omega$=angular speed

Put the value into the formula

$L=6.42\times10^{23}\times(2.28\times10^{11})^2\times\dfrac{2\pi}{687\times24\times3600}$

$L=3.53\times10^{39}\ kg-m^2/s$

Hence, The magnitude of the angular momentum of the Mars in a circular orbit around the sun is $3.53\times10^{39}\ kg-m^2/s$