A planet is revolving around the sun in an elliptical orbit. The mass of planet is m, angular momentum of planet about sun is L, and length of semi major axis is a and eccentricity is e. Time period of revolution of planet is given by

Option 1 -

T = (2πma²√1+e²)/L

Option 2 -

T = (πma²√1-e²)/L

Option 3 -

T = (2πma²√1-e²)/L

Option 4 -

T = (πma²√1-e²)/3L

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 weeks ago

Correct Option - 3

Detailed Solution:

b = a√1 - e²
dA/dt = L/2m
T = A / (L/2m)
T = 2mA/L = 2mπab/L = 2mπa²√1 - e²/L

0 Upvotes 0 Downvotes

Similar Questions for you

A wire can be broken by a load of 20 kg-wt. The force required to break wire of same material with twice the diameter will be

alok kumar singh

The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If speed of satellite A is 2V, then the speed of satellite B will be

alok kumar singh

At a point away from planet of radius 6400 km, the gravitational potential and field are –6.4 × 10⁷ SI units and 6.4 SI units respectively. Find height of that point above surface of planet.

alok kumar singh

$\frac{G M}{r} = 6.4 * 1 0^{7}$ . (i)

$\frac{G M}{r^{2}} = 6.4$ . (ii)

$r = \frac{6.4 * 1 0^{7}}{6.4}$

= 10⁷ m

= 10,000 km

R + h = 10,000

h = 10,000 – 6400 = 3600 km

Mass of moon is $\frac{1}{81}$ times the mass of a planet and radius is $\frac{1}{9}$ times the radius of the planet. The ratio of escape speed from planet to escape speed from moon is ________.

alok kumar singh

-> $V_{e s c} = \sqrt[]{\frac{2 G M}{R}}$

$R a t i o = \sqrt[]{\frac{81}{9}} = 3$

If the gravitational acceleration at the surface of earth is g, then increase in potential energy in lifting an object of mass m from earth surface to a height equal to half of radius of earth from surface will be:

alok kumar singh