The ratio of Earth’s orbital angular momentum (about$$ the $$sun) to its mass is $$4.4$$×$$1015$$ $$m2/s$$. The area enclosed by the earth’s orbit is N×$$1023$$ $$m2$$. Find the value of N ?

Question

The ratio of Earth’s orbital angular momentum (about$$ the $$sun) to its mass is  $$4.4$$×$$1015$$ $$m2/s$$. The area enclosed by the earth’s orbit is  N×$$1023$$ $$m2$$. Find the value of N ?

Infinity Learn · Accepted Answer

GivenAngular momentum per mass $$LM=4.4$$×$$1015$$ $$m2/sBy$$ Kepler’s second law we use Areal velocity $$=$$  Area SweptTime for one Revolution of Earth about the sun Also we use Areal velocity $$=$$  $$L2M$$⇒ Area Swept $$=$$  $$L2M$$ (Time$$ for one Revolution of Earth about the $$sun) ⇒ Area Swept $$=$$  $$124.4$$×$$1015365$$×$$24$$×$$60$$×$$60$$⇒ Area Swept $$=$$  $$0.69$$×$$1023$$ $$m2$$

Gravitational force and laws

Remember concepts with our Masterclasses.

The ratio of Earth’s orbital angular momentum (about the sun) to its mass is $4.4 \times 10^{15} m^{2} / s$ . The area enclosed by the earth’s orbit is $N \times 10^{23} m^{2}$ . Find the value of N ?

Ready to Test Your Skills?

free study material

Gravitational force and laws

Remember concepts with our Masterclasses.

The ratio of Earth’s orbital angular momentum (about the sun) to its mass is 4.4×1015 m2/s. The area enclosed by the earth’s orbit is N×1023 m2. Find the value of N ?

Ready to Test Your Skills?

free study material

The ratio of Earth’s orbital angular momentum (about the sun) to its mass is $4.4 \times 10^{15} m^{2} / s$ . The area enclosed by the earth’s orbit is $N \times 10^{23} m^{2}$ . Find the value of N ?