JEE Advanced (Single Correct MCQs): Gravitation - JEE MCQ

 Initial potential energy of the system.

 Final P.E. of the system.

According to Kepler's law 

Note :  A satellite revolving near the earth's surface has a time period of 84.6 min.
We know that as the height increases, the time period increases. Thus the time period of the spy satellite should be slightly greater than 84.6 minutes.

∴ T_s = 2 hr

The components of acceleration are as shown

The resultant of transverse and radial component of the acceleration is represented 

Note : The gravitational force of attraction between the stars will provide the necessary centripetal forces.
In this case angular velocity of both stars is the same.
Therefore time period remains the same. 

For r > R

Force on the test mass m is F = m × | E_g |

Where E_g is the gravitational field intensity at the point of observation

 where M is the total mass of the
spherical system.

Let us consider a circular elemental area of radius x and thickness dx. The area of the shaded portion = 2πxdx.
Let dm be the mass of the shaded portion.

The gravitational potential of the mass dm at P is

Suppose 16R² + x² = t²
⇒  2xdx = 2tdt ⇒ xdx = tdt
Also for x = 3R, t = 5R
and for x = 4R, 

On integrating equation (1),  taking the above limits, we get

V is the orbital velocity. If V_C is the escape velocity then   The kinetic energy at the time of ejection

The mass of the wire = 10^–3 × 1.2 × 10⁵ = 120 kg
Let g_pM be the acceleration due to gravity at point M which is the mid point of the wire and is at a depth of 

Let gp be the acceleration due to gravity at the surface of the planet.

and 

∴ Force = mass of wire × g_pM = 120 × 0.9 = 108 N

KEY CONCEPT :  The centripetal force is provided by the gravitational force of attraction mRω² = GMωR^–5/2

The gravitational field intensity at the point O is zero (as the cavities are symmetrical with respect to O). Now the force acting on a test mass m₀ placed at O is given by

F = m₀ E = m₀ × 0 = 0

Now, y² + z² = 36 represents the equation of a circle with centre (0, 0, 0) and radius 6 units the plane of the circle is perpendicular to x-axis.

Note :  Since the spherical mass distribution behaves as if the whole mass is at its centre (for a point outside on the sphere) and since all the points on the circle is equidistant from the centre of the sphere, the circle is a gravitational equipotential.
The same logic holds good for option (d).

For r > R, the gravitational field is 

For r < R, the gravitational field is 

Force on satellite is always towards earth, therefore, acceleration of satellite S is always directed towards centre of the earth. Net torque of this gravitational force F about centre of earth is zero. Therefore, angular momentum (both in magnitude and direction) of S about centre of earth is constant throughout. Since the force F is conservative in nature, therefore mechanical energy of satellite remains constant. Speed of S is maximum when it is nearest to earth and minimum when it is farthest.

Let the mass of P be m.

∴ The mass of R = 9 m
If the radius of P = r
Then the radius of Q = 2r

and radius of R = 9^1/3r

The potential energy is a combined property of the three mass system. The kinetic energy of mass m is only its energy which decreases as it moves. (b) is the correct option.