Question

the earth is a point mass of 6 x 1024 kg revolving around the sun at a distance of 1.
kg revolving around the sun at a distance of 1.5 x 108 km and in
of T 3.14 x 10 seconds, then the angular momentum of the earth around the s
time of T= 3.14
(A) 1.2 x 101 kg-m/s (B) 1.8 x 1020 kg-m/s (C) 1.5 x 1037 ko-m/s (D) 2.7 x 1040 kg-m/s​

Answer 1

Answer:

Here is the answer to your question

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Explanation:

Answer 2

The angular momentum of the earth around the sun is $2.7\times10^{40}\ kg-m/s$

(D) is correct option

Explanation:

Given that,

Mass $M=6\times10^{24}\ kg$

Distance $r =1.5\times10^{8}\ km$

Suppose time is $T= 3.14 \times10^{7}\ sec$

We need to calculate the angular momentum of the earth around the sun

Using formula of angular momentum

$L=I\omega$

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

Put the value into the formula

$L=6\times10^{24}\times(1.5\times10^{8}\times10^{3})^2\times\dfrac{2\pi}{3.14 \times10^{7}}$

$L=2.7\times10^{40}\ kg-m/s$

Hence, The angular momentum of the earth around the sun is $2.7\times10^{40}\ kg-m/s$

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Topic :

https://brainly.in/question/10725445