Gravitation Test - 9

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​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 rg=Gm/a² for r

• The orbital velocity of the satellite depends on its altitude above Earth.
• The nearer Earth, the faster the required orbital velocity. At an altitude of 200 kilometers, the required orbital velocity is just over 27,400 kph.
• To maintain an orbit that is 35,786 km above Earth, the satellite must orbit at a speed of about 11,300 kph. That orbital speed and distance permits the satellite to make one revolution in 24 hours.
• To achieve his speed, Judging from the mass of fuel needed, it stands to reason that you'd use a more powerful launch vehicle with a bi-propellant upper stage to get the satellite to GEO.

• The gravitational force acting by a spherically symmetric shell upon a point mass inside it, is the vector sum of gravitational forces acted by each part of the shell, and this vector sum is equal to zero.
• That is, a mass m within a spherically symmetric shell of mass M will feel no net force.

Time period of revolution of earth around sun T_e= 1 Year

Time period of revolution of planet around sun,T_p=0.5 Year

Orbital size of earth, r_e= 1 A.U

Orbital size of the planet,r_p=?

Applying Kepler's third law we get: 

• Let L be the width of the film. Since the film has two surfaces, the total length along which the surface force acts on the slider is 2L.
• The surface tension S in the film is defined as the ratio of the surface force F to the length d (perpendicular to the force) along which the force acts S=F/d
• Hence, in the case, d = 2L S=F/2L

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.

Tides refer to the rise and fall of our oceans' surfaces. It is caused by the attractive forces of the Moon and Sun's gravitational fields as well as the centrifugal force due to the Earth's spin.