Home
Class 11
PHYSICS
The ratio of the earth's orbital angular...

The ratio of the earth's orbital angular momentum (about the Sun) to its mass is `4.4 xx 10^(15) m^(2) s^(-1)`. The area enclosed by the earth's orbit is approximately-___________.

Text Solution

Verified by Experts

The correct Answer is:
NA

Areal velocity of a planet around the sun is constant and is given by
`(dt)/(dt)=L/(2m)impliesdA=L/(2m)dt`
Integrating both sides `int dA=L/(2m)intdtimpliesA=L/(2m)/T`
where `L=` angular momentum of the planet (earth abut the sun and `m=` mass of planet (earth).
Hence `A=L/(2m)T=1/2xx4.4xx10^(15)xx365xx24xx3600m^(2)`
Area `=6.94xx10^(22)m^(2)`
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    CENGAGE PHYSICS|Exercise True/False|1 Videos
  • GRAVITATION

    CENGAGE PHYSICS|Exercise SCQ_TYPE|12 Videos
  • GRAVITATION

    CENGAGE PHYSICS|Exercise Integer|5 Videos
  • FLUID MECHANICS

    CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos
  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Integer|9 Videos

Similar Questions

Explore conceptually related problems

The orbital angular momentum of an electron in 2s orbital is

The orbital angular momentum of an electron is 2s orbital is

The orbital angular momentum of an electron in 2s -orbital is

The orbital angular momentum of an electron in 2s orbital is ………

The orbital angular momentum of electron in 4s orbital of H atom is ……….

The ratio of the angular momentum of the orbital electron in the first orbit to that in the 2nd orbital is

The mean radius of the earth's orbit around the sun is 1.5 xx 10^(11)m and that of the orbit of mercury is 6 xx 10^(10)m . The mercury will revolve around the sun is nearly