A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is ---------. 'P' has the time period of 24 hours.

Option 1 -

6√2

Option 2 -

Option 3 -

Option 4 -

6/√2

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

(T / 24) = (3/12)^ (3/2) ⇒ T = 3 hours

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