Test: Gravitation - 1 - JEE MCQ

Gravity does not depend upon shape and size of the celestial body. It depends only on the mass of the two bodies and distance between them.
Fold = GmM/R² and Fnew = GmM/R²

Total time period will be 84.6 min when ball released from one end and it come backs to same point as in oscillation.
When ball is released from one end then time taken to reach other end will be half of total time period then that will be 42.3 min.
 

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​

Applying the conservation of momentum, we get:
V = velocity of 5M, V1 = velocity of 4M, V2 = velocity of M and ATQ V2 = -V
5MV = 4MV1 + M(-V)
6MV = 4MV1
V1 = 3/2 V
Now, V_o = orbital velocity and V_e = escape velocity
V^e = √2 V_o 
In a bound orbit the object is gravitationally bound to the body that is the source of gravity (like the Earth is bound to the Sun, or the Moon to the Earth). An unbound orbit is typically hyperbolic, and the object will escape from the source of gravity
 

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​

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.

Answer :- a

Solution :- L = mvr

⇒L=m√GMr/r 

L = m√GMr−−−−.....(1)

L = 2mdA/dt....(2)

From (1) and (2)

dA/dt ∝ (r)^1/2

= √(dA/dt)1/(dA/dt)2

= √4/1 = 2/1

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

As g is inversely proportional to (R+h)2 when we go away from earth's surface g is inversely proportional to square of the distance and g is directly proportional to (R-d) when we go inside the surface of Earth therefore g is directly proportional to distance travelled inside the Earth
g=Gmr/a³=r for r<R
g=Gm/a² for r<R

The total energy of the system remains constant.

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.

Minimum colatitude=sin θ=R/R+h
θ=sin^-1(R/R+h)
The curved area AB on earth =2πR²(1−sinθ)
Area on earth escaped from satellite =4πR²−2πR²(1−sinθ)
=2πR²(1+sinθ)

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.

Communication satellites are supposed to be geostationary in nature. So, from the above properties options B is correct.

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.

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.

The velocity of satellite V=√GM​/R​⇒V∝1/√R​​. 
The time period of satellite is given by  T=2π√ R³/GM​​⇒T∝R^1.5
The total energy is given by  T.E.=−GMm/2R​⇒T.E.∝−1/M​ 
Here, G is constant, R is orbital radius, M is mass of earth and m is mass of satellite.
Near the earth surface, R has the least value and therefore time period and total energy are minimum while velocity is maximum. So, options A, B, C are correct.