Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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The document Irodov Solutions: Universal Gravitation- 1 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 200. A planet of mass M moves along a circle around the Sun with velocity v = 34.9 km/s (relative to the heliocentric reference frame). Find the period of revolution of this planet around the Sun.

Ans. We have

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Thus   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(Here ms the mass of the Sun.)

Irodov Solutions: Universal Gravitation- 1 Notes | EduRevIrodov Solutions: Universal Gravitation- 1 Notes | EduRev

(The answer is incorrectly written in terms of the planetary mass M)


Q. 201.  The Jupiter's period of revolution around the Sun is 12 times that of the Earth. Assuming the planetary orbits to be circular, find:
 (a) how many times the distance between the Jupiter and the Sun exceeds that between the Earth and the Sun;
 (b) the velocity and the acceleration of Jupiter in the heliocentric reference frame. 

Ans. For any planet

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(a) Thus   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(b)    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

where T = 12 years. ms - mass of ths Sun.
Putting the values we get V= 12.97 km/s 

Acceleration  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 202. A planet of mass M moves around the Sun along an ellipse so that its minimum distance from the Sun is equal to r and the maximum distance to R. Making use of Kepler's laws, find its period of revolution around the Sun. 

Ans. Semi-major axis= (r +/R)/2 

It is sufficient to consider the motion be along a circle o f semi-m ajor axis r + R/2 for T does not depend on eccentricity.

Hence  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(again ms is the mass of the Sun)


Q. 203. A small body starts falling onto the Sun from a distance equal to the radius of the Earth's orbit. The initial velocity of the body is equal to zero in the heliocentric reference frame. Making use of Kepler's laws, find how long the body will be falling. 

Ans. We can think of the body as moving in a very elongated orbit of maximum distance R and minimum distance 0 so semi major axis = R/2. Hence if τ is the time of fall then

or    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 204. Suppose we have made a model of the Solar system scaled down in the ratio η but of materials of the same mean density as the actual materials of the planets and the Sun. How will the orbital periods of revolution of planetary models change in this case?

Ans. Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

If the distances are scaled down, R3/2 decreases by a factor η3/2 and so does ms . Hence T does not change.


Q. 205. A double star is a system of two stars moving around the centre of inertia of the system due to gravitation. Find the distance between the components of the double star, if its total mass equals M and the period of revolution T. 

Ans. The double star can be replaced by a single star of mass   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev moving about the centre of mass subjected to the force

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

or,  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 206. Find the potential energy of the gravitational interaction
 (a) of two mass points of masses m1 and m2  located at a distance r from each other;
 (b) of a mass point of mass m and a thin uniform rod of mass M and length l, if they are located along a straight line at a distance a from each other; also find the force of their interaction

Ans. (a) The gravitational potential due to m1 at the point of location of m2 :

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Similarly   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Hence  

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(b) Choose the location of the point mass as the origin. Then the potential erfeigy dU of an element of mass  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev of the rod in the field of the point mass is

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

where x is the distance between the element and the point (Note that the rod and the point mass are on a straight line.) If then a is the distance of the nearer end of the rod from the point mass.

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

The force of interaction is 

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Minus sign means attraction.


Q. 207. A planet of mass m moves along an ellipse around the Sun so that its maximum and minimum distances from the Sun are equal to r and r2  respectively. Find the angular momentum M of this planet relative to the centre of the Sun. 

Ans. As the planet is under central force (gravitational interaction), its angular momentum is conserved about the Sun (which is situated at one of the focii of the ellipse)

So,    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev       (1)

From the conservation of mechanical energy of the system (Sun + planet),

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

or,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Thus,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev     (2)

Hence   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 208. Using the conservation laws, demonstrate that the total mechanical energy of a planet of mass m moving around the Sun along an ellipse depends only on its semi-major axis a. Find this energy as a function of a. 

Ans. From the previous problem, if r1 , r2 are the maximum and minimum distances from the sun to the planet and v1 , v2 are the corresponding velocities, then, say,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

where 2d = major axis = r1 + r2. The same result can also be obtained directly by writing an equation analogous to Eq (1) of problem 1.191.

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(Here Af is angular momentum of the planet and m is its mass). For extreme position Irodov Solutions: Universal Gravitation- 1 Notes | EduRev and we get the quadratic

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

The sum of the two roots of this equation are 

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Thus   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 209. A planet A moves along an elliptical orbit around the Sun. At the moment when it was at the distance r0  from the Sun its velocity 'was equal to v0 and the angle between the radius vector r0  and the velocity vector v0 was equal to α. Find the maximum and minimum distances that will separate this planet from the Sun during its orbital motion. 

Ans. From the conservtion of angular momentum about the Sun.

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev        (1)

From conservation of mechanical eneigy,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev
or,    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev
or,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev
So,    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev
Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

where   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 210. A cosmic body A moves to the Sun with velocity v0(when far from the Sun) and aiming parameter l the arm of the vector v0 elative to the centre of the Sun (Fig. 1.51). Find the minimum distance by which this body will get to the Sun. 

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Ans. At the minimum separation with the Sun, the cosmic body’s velocity is perpendicular to its position vector relative to the Sun. If rmin be the sought minimum distance, from conservation of angular momentum about the Sun (C).

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev   (1)

From conservation of mechanical energy of the system (sun + cosmic body),

  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

or,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So,  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Hence, taking positive root

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 211. A particle of mass in is located outside a uniform sphere of mass M at a distance r from its centre. Find:
 (a) the potential energy of gravitational interaction of the particle and the sphere;
 (b) the gravitational force which the sphere exerts on the particle. 

Ans. Suppose that the sphere has a radius equal to a. We may imagine that the sphere is made up of concentric thin spherical shells (layers) with radii ranging from 0 to a, and each spherical layer is made up of elementry bands (rings). Let us first calculate potential due to an elementry band of a spherical layer at the point of location of the point mass m (say point P) (Fig.). As all the points of the band are located at the distance l from the point P, so,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev    (1)

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Differentiating Eq. (3), we get

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev      (4)

Hence using above equations

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev   (5)

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Now integrating this Eq. over the whole spherical layer

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev      (6)

Equation (6) demonstrates that the potential produced by a thin uniform spherical layer outside the layer is such as if the whole mass of the layer were concentrated at it’s centre; Hence the potential due to the sphere at point P;

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev     (7)

This expression is similar to that of Eq. (6)

Hence thte sought potential eneigy of gravitational interaction of the particle m and the sphere,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(b) Using the Eq.,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

So    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev      (8)


Q. 212. Demonstrate that the gravitational force acting on a par- ticle A inside a uniform spherical layer of matter is equal to zero. 

Ans. (The problem has already a dear hint in the answer sheet of the problem book). Here we adopt a different method.
Let m be the mass of the spherical layer, wich is imagined to be made up of rings. At a point inside the spherical layer at distance r from the centre, the gravitational potential due to a ring element of radius a equals,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Hence    Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Hence gravitational field strength as well as field force becomes zero, inside a thin sphereical layer.


Q. 213. A particle of mass m was transferred from the centre of the base of a uniform hemisphere of mass M and radius R into infinity. What work was performed in the process by the gravitational force exerted on the particle by the hemisphere? 

Ans. One can imagine that the uniform hemisphere is made up of thin hemispherical layers of radii ranging from 0 to R. Let us consider such a layer (Fig.). Potential at point O, due to this layer is,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(This is because all points of each hemispherical shell are equidistant from O.)

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Hence, the work done by the gravitational field force on the particle of mass m, to remove it to infinity is given by the formula

A = mφ , since φ = 0 at infinity.

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Hence the sought work,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(The work done by the external agent is - A.)


Q. 214. There is a uniform sphere of mass M and radius R. Find the strength G and the potential φ of the gravitational field of this sphere as a function of the distance r from its centre (with r < R and r > R). Draw the approximate plots of the functions G (r) and φ (r). 

Ans. In the solution of problem 1.211, we have obtained φ and G due to a uniform shpere, at a distance r from it’s centre outside it We have from Eqs. (7) and (8) of 1.211,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev    (A)

Accordance with the Eq. (1) of the solution of 1.212, potential due to a spherical shell of radius a, at any point, inside it becomes

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev   (B)

For a point (say P) which lies inside the uniform solid sphere, the potential φ at that point may be represented as a sum.

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

where φ1 is the potential o f a solid sphere having radius r and φ2 is the potential of the layer of radii r and R. In accordance with equation (A)

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

The potential φ2 produced by the layer (thick shell) is the same at all points inside it. The potential φ2 is easiest to calculate, for the point positioned at the layer’s centre. Using Eq. (B)

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

where   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

is the mass of a thin layer between the radii r and r + dr.

Thus   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

From the Eq.   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

or   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(where  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev is the density of the sphere)     (D)

The plots φ (r) and G (r) for a uniform sphere of radius R are shown in figure of answersheet.

Alternate : Like Gauss’s theorem of electrostatics, one can derive Gauss’s theorem for gravitation in the form 

Irodov Solutions: Universal Gravitation- 1 Notes | EduRevat a point  inside the sphere at a distance r from its centre, let us consider a Gaussian surface of radius r, Then,

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Integrating and summing up, we get

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

And from Gauss’s theorem for outside it :

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Thus   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 215. Inside a uniform sphere of density p there is a spherical cavity whose centre is at a distance l from the centre of the sphere. Find the strength G of the gravitational field inside the cavity. 

Ans. Treating the cavity as negative mass of density - p in a uniform sphere density + p and using the superposition principle, the sought field strength is :

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(where  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev are the position vectors of an orbitrary point P inside the cavity with respect to centre of sphere and cavity respectively.)

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Q. 216. A uniform sphere has a mass M and radius R. Find the pressure p inside the sphere, caused by gravitational compression, as a function of the distance r from its centre. Evaluate p at the centre of the Earth, assuming it to be a uniform sphere. 

Ans. We partition the solid sphere into thin spherical layers and consider a layer of thickness dr lying at a distance r from the centre of the ball. Each spherical layer presses on the layers within it The considered layer is attracted to the part of the sphere lying within it (the outer part does not act on the layer). Hence for the considered layer

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(The pressure.must vanish at r = R.)

or,   Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Putting r = 0, we have the pressure at sphere’s centre, and treating it as the Earth wheremean density is equal to  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

we have , Irodov Solutions: Universal Gravitation- 1 Notes | EduRev


Q. 217. Find the proper potential energy of gravitational interaction of matter forming
 (a) a thin uniform spherical layer of mass m and radius R;
 (b) a uniform sphere of mass m and radius R (make use of the answer to Problem 1.214). 

Ans. (a) Since the potential at each point of a spherical surface (shell) is constant and is equal to  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev  [as we have in Eq. (1) of solution of problem 1.212]

We obtain in accordance with the equation

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

(The factor 1/2 is needed otherwise contribution of different mass elements is counted twice.]

(b) In this case the potential inside the sphete depends only on r (see Eq. (C) of the solution of oroblem 1.214)

  Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Here dm is the mass of an elementry spherical layer confined between the radii r and r + dr :

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev
Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

After integrating, we get 

Irodov Solutions: Universal Gravitation- 1 Notes | EduRev

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