Gravitation

Ans1

Orbital velocity is √GM/R so the ratio will be 1:1

@saurabhtiwari I don't think you are correct as in this case Newton's inverse square law is not followed

Orbital velocity Vo=(GM/R)½
For satellite of mass m and radius R Vo1= (GM/R)½.......(1)
For satellite of mass 2m and radius 2R Vo2= (2GM/2R)½ =>(GM/R)½.............(2)From equation 1/2
Vo1/Vo2 = 1:1

@siddharthaghosh see the gravitational force and centripetal force need to balance each other
In this case the gravitational force is inversely proportional to r contarary to r^2.
Thus MV^2/R. =K*M/R (K is some constant and yes the force has to depend on mass)
Thus you can clearly see mass and radius both cancel out and orbital velocity ration us 1:1

@pranaysinha why is mv^2/r = k m/r
Wont it be mv^2/r = gm(1) m(2)/r^2
Then ans is not matchingUsing V orb =rootGM/r ans matching

@siddharthaghosh ans should be 4th option as formula root GM/R is correct for speed of orbit but here M is mass of earth not satellite which is constant.
Orbits have defined speed which don't depend on mass of satellite.

@siddharthaghosh
@PranaySingh
When you say force has to depend on mass then yes it should depend on mass of both the bodies and not only one

@vedantzope force depends on mass of body but orbital speed does not. Satellite is in free fall. U have used wrong formula of force and the masses that cancel out are satellite masses so they don't have significance in orbital speed.

@siddharthaghosh in my equation k is nothing but G m1 , I just wrote k to signify a constant which does not matter my calculations

@pranaysingh even I did the same and I agree orbital velocity does not depend on mass of sattelite but according to deriviation as per the relation given in question orbital velocity is coming as root GM where M is mass of planet

@vedantzope but F=Gm1m2/r^2 why Gm1m2/r = mv^2/r .
Pls explain that part

@siddharthaghosh and so one r factor remain.

@siddharthaghosh @PranaySingh
Because they have given force is inversely proportional to distance from centre
This is some imaginary planet so necessarily newtons law is not followed here , as while deriving orbital velocity we equate force to mv^2/r , I have done the same so here the arrival velocity nor depends on radius and neither on mass as before

@vedantzope o now got it
Thanks

So the speed is becoming independent of mass for this hypothetical case

@vedantzope oo lol didn't see it. Good one bro!