Test: Gravitation (October 4) - NEET MCQ

Let’s see the formulae for each quantity given there,
Gravitational potential energy  =−GMm/r​
angular velocity =(GM/r³​)^0.5
linear orbital velocity  =(GM/r​)^0.5
centripetal acceleration  =GM/ r²
We can observe that, the quantities in the options B,C,D are all indirectly proportional to r
i.e. ω, v, a α(1/rⁿ)​ where n is a distinct positive integer in each quantity.
So, with increase in r all these quantities decrease.
Now, (-potential energy) α(1/r)​ i.e. negative of potential energy decreases with increase in r. Therefore, potential increases with increase in orbital radius.

If a planet is orbiting the sun then |U|> |K| else it will leave the system because it won’t be under the influence of the sun’s gravity.

The gravitational field inside a uniform spherical shell is 0 from gauss law for gravitation since no mass is enclosed in any Gaussian surface.
Since gravitational potential is given by φ=−∫ gdr, hence, φ=constant since g=0.
Since the gravitational field is 0 everywhere, it is apparently the same everywhere.
Answer is B,C,D.

The total energy of the system remains constant.

Here, the angular momentum is conserved i.e. mvr is constant. Where r is the distance from centre of the sun to centre of the planet.
Now, consider one point each from both the mentioned paths. (Shown in fig.)
Applying conservation of angular momentum for these points we get 
mv₁​r₁​=mv₂​r₂​. Simplifying this, v₁/​v₂​​=r₂​/r₁​​<1
Time period are given by 
t_1​=L/ v₁​, t₂​= L​/ v₂
Comparing them by taking the ratio 
t₁/t_2​​=(L​/ v₁)×(v₂/L)
So, t₁/t₂=v₂​/v₁​​>1.Thus t_1​>t₂​

Due to air drag some mechanical energy of satellite will converted into heat energy, there will be loss of ME of satellite, so radius of orbit will decrease and satellite will follow a spiral path towards earth. 
If r is decreases,
P.E.=−GMm​/r, P.E. is also decreases.
K.E=GMm​/2r K.E increasing, increased speed in spite of decrease in M.E. But rate of P.E decreases is more than the rate at which M.E decreases.
Angular momentum L=mrv=mr√GM​r=m√GMr​⇒L∝√r​.
So, angular momentum decreases as r decreases.

As the gravitational force on satellites due to earth acts always towards the centre of earth, thus acceleration of S is always directed towards the centre of the earth.
Also, as there is no external force so according to conservation of energy, total mechanical energy of S is constant always.
Also, as in the absence of external torque L is constant in magnitude and direction.
Thus, mrv=constant ⟹v varies as r changes
Hence, p=mv is not constant.

Kinetic Energy of the Satellite =GMm​/2r
Potential Energy of the satellite =−GMm​/r 
Total Energy of the Satellite =−GMm​/2r
Hence, Kinetic Energy Cure will be above 0 , hence X 
Potential energy will be the lowest curve , hence Y 
Total Energy is the curve is below 0 , but higher magnitude than Potential Energy  . Hence, Z

Gravitational force by mass M on another body of m at a distance of r is given by F=GMmr²
For given figure Right diagonal is x axis and left y axis in figure A forces due to 5M will cancel out each other so effective force will be due to only 3M an M. In figure B 2M forces will cancel out each other and effective force will be due to 3M and M.
Hence total force on M in both the cases will be the same.
Similarly, in figure C, 5M mass will cancel each other out and in Figure D, 2M will cancel each other out and effective force will be due to 3M and M on the mass of 2M. So, force on c and d will be the same.
Force on C will be more than B.
Hence answer is F_A​=F_B​<F_C​=F_D​