Mathematics KEPLER’S LAWS AND UNIVERSAL LAW OF GRAVITATION

### Topic covered

star INTRODUCTION
star KEPLER’S LAWS
star UNIVERSAL LAW OF GRAVITATION

### INTRODUCTION

color{green} ✍️  Gravity is one of the most basic forces in the universe. It plays a fundamental role not only in the structure of our solar system but also in the way objects behave on Earth.

color{green} ✍️ Gravitation is the force of attraction between two objects in the universe. Gravitation may be the attraction of objects by the earth.

color{green} ✍️  So The gravitational force is a force that attracts any objects with mass. You, right now, are pulling on every other object in the entire universe! This is called Newton's Universal Law of Gravitation.

color{green} ✍️  So here we start from Kepler's laws of planetary motion, three scientific laws describing the motion of planets around the Sun.

### KEPLER’S LAWS

color{green} ✍️ The three laws of Kepler can be stated as follows:

 colorbbul{purple} {1. "Law of orbits :"}

● "All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse (Fig. 8.1a)"..
● This law was a deviation from the Copernican model which allowed only circular orbits. The ellipse, of which the circle is a special case, is a closed curve which can be drawn very simply as follows.

=> Select two points F_1 and F_2. Take a length of a string and fix its ends at F_1 and F_2 by pins.
=> With the tip of a pencil stretch the string taut and then draw a curve by moving the pencil keeping the string taut throughout.(Fig. 8.1(b))
=> The closed curve you get is called an ellipse. Clearly for any point T on the ellipse, the sum of the distances from F_1 and F_2 is a constant. F_1, F_2 are called the focii.
=> Join the points F_1 and F_2 and extend the line to intersect the ellipse at points P and A as shown in Fig. 8.1(b). The midpoint of the line PA is the centre of the ellipse O and the length PO = AO is called the semimajor axis of the ellipse. For a circle, the two focii merge into one and the semi-major axis becomes the radius of the circle.

 color{purple}bbul{2. "Law of areas :"}

● "The line that joins any planet to the sun sweeps equal areas in equal intervals of time" (Fig. 8.2).
● This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.

 color{purple}bbul{3. "Law of periods :"}
● "The square of the time period of revolution of a planet is proportional to the cube "
"of the semi-major axis of the ellipse traced out by the planet."
● The table below gives the approximate time periods of revolution of nine planets around the sun along with values of their semi-major axes.

 color{orange}{"Table " 1}  color{green} {"Data from measurement of planetary motions given below confirm Kepler’s Law of Periods"}

color{blue}{a ≡ "Semi-major axis in units of "10^10 "m"}.
color{blue}{T ≡" Time period of revolution of the planet in years " (y).
color{blue}{Q ≡" The quotient "( T^2//a^3 ) " in units of "10^(-34) y^2 m^(-3).

color{green}{ :-☞ } The law of areas can be understood as a consequence of conservation of angular momentum which is valid for any central force .
● A central force is such that the force on the planet is along the vector joining the sun and the planet.
● Let the sun be at the origin and let the position and momentum of the planet be denoted by r and p respectively. Then the area swept out by the planet of mass m in time interval Δt is (Fig. 8.2) ΔA given by

ΔA=1//2 (r xx v Δt)

Hence

ΔA//Δt=1//2 (r xx p)//m, (since v=p//m)

color{blue}{=L//(2m)}

=> where v is the velocity,  L is the angular momentum equal to ( r xx p ). For a central force, which is directed along r, L is a constant as the planet goes around.
=> Hence, Δ A //Δt is a constant according to the last equation. This is the law of areas. Gravitation is a central force and hence the law of areas follows.
Q 3210401319

Let the speed of the planet at the perihelion P in Fig. 8.1(a) be vP and the Sun-planet distance SP be r_P. Relate {r_P, v_P} to the corresponding quantities at the aphelion {r_A, v_A}. Will the planet take equal times to traverse BAC and CPB ?

Solution:

The magnitude of the angular momentum at P is color{green} {L_p = m_p r_p v_p}, since inspection tells us that r_p and v_p are mutually
perpendicular. Similarly, color{purple} {L_A = m_p r_A v_A}.

From angular momentum conservation

color{orange} {m_p r_p v_p = m_p r_A v_A}

or (v_p)/(v_A)=(r_A)/(r_p)

Since

color{blue}{r_A > r_p, v_p > v_A}

.

The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in Fig. 8.1. From Kepler’s second law, equal areas are swept in equal times. Hence the planet will take a longer time to traverse BAC than  CPB.

### UNIVERSAL LAW OF GRAVITATION

color{green} ✍️ Legend has it that observing an apple falling from a tree, Newton was inspired to arrive at an universal law of gravitation that led to an explanation of terrestrial gravitation as well as of Kepler’s laws. Newton’s reasoning was that the moon revolving in an orbit of radius R_m was subject to a centripetal acceleration due to earth’s gravity of magnitude

color{blue}{a_m=(V^2)/(R_m) =(4 pi^2R_m)/(T^2)..........................(8.3)}

=> where V is the speed of the moon related to the time period T by the relation  V =2 π R //T.
☞ The time period T is about 27.3 days and R_m was already known then to be about 3.84 xx 10^8 m. If we substitute these numbers in equation (8.3), we get a value of am much smaller than the value of acceleration due to gravity g on the surface of the earth, arising also due to earth’s gravitational attraction.

☞ This clearly shows that the force due to earth’s gravity decreases with distance. If one assumes that the gravitational force due to the earth decreases in proportion to the inverse square of the distance from the center of the earth, we will have in agreement with a value of g ; 9.8 m s^(-2) and the value of am from Eq. (8.3).

color{brown}{a_m α R_m^(−2) : g α R_E^(-2)}

and we get color{blue}{g/(a_m) =(R_m^2)/(R_E^2) ; 3600.......................(8.4)}

These observations led Newton to propose the following Universal Law of Gravitation : Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

● The quotation is essentially from Newton’s famous treatise called ‘Mathematical Principles of Natural Philosophy’ (Principia for short). Stated Mathematically, Newton’s gravitation law reads : The force F on a point mass m_2 due to another point mass m_1 has the magnitude

color{blue}{|F|= G(m_1m_2)/(r^2)..........................(8.5)}

● Equation (8.5) can be expressed in vector form as

F=G (m_1m_2)/r^2 (-hat r)=-G (m_1m_2)/(r^2) hat r

color{blue}{ :. F=-G (m_1m_2)/(|r^3|) hat r}

=> where G is the universal gravitational constant, hat r is the unit vector from m_1 to m_2 and r=r_2-r_1 as shown in Fig. 8.3.

● The gravitational force is attractive, i.e., the force F is along – r. The force on point mass m_1 due to m_2 is of course – F by Newton’s third law.
● Thus, the gravitational force F_(12) on the body 1 due to 2 and F_(21) on the body 2 due to 1 are related as F_(12) = – F_(21). Before we can apply Eq. (8.5) to objects under consideration, we have to be careful since the law refers to point masses whereas we deal with extended objects which have finite size.
● If we have a collection of point masses,the force on any one of them is the vector sum of the gravitational forces exerted by the other point masses as shown in Fig 8.4.

● The total force on m_1 is

color{blue}{F_1=(Gm_2m_1)/(r_(21^2)) hat r_(21) + (G m_3m_1)/(r_(31)^2) hat r _(31) + (Gm_4m_1)/(r_(41)^2) hat r_(41)}

● For the gravitational force between an extended object (like the earth) and a point mass, Eq. (8.5) is not directly applicable. Each point mass in the extended object will exert a force on the given point mass and these force will not all be in the same direction.
● We have to add up these forces vectorially for all the point masses in the extended object to get the total force. This is easily done using calculus. For two special cases, a simple law results when you do that :

color{green}{(1) " The force of attraction between a hollow spherical shell of uniform density and a "}
color{green}{"point mass situated outside is just as if the entire mass of the shell is concentrated at "}
color{green}{"the centre of the shell"}. Gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction prependicular to this line. The components prependicular to this line cancel out when summing over all regions of the shell leaving only a resultant force along the line joining the point to the centre. The magnitude of this force works out to be as stated above.

color{green}{(2) " The force of attraction due to a hollow spherical shell of uniform density, on a"}
color{green}{ "point mass situated inside it is zero"}. Qualitatively, we can again understand this result. Various regions of the spherical shell attract the point mass inside it in various directions. These forces cancel each other completely.

Q 3220601511

Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC.

(a) What is the force acting on a mass 2m placed at the centroid G of the triangle?
(b) What is the force if the mass at the vertex A is doubled ?

Gravitational force on m Take AG = BG = CG = 1 m (see Fig. 8.5)

Solution:

(a) The angle between GC and the positive x-axis is 30° and so is the angle between GB and the negative x-axis. The individual forces
in vector notation are

color{green} {F_(GA)= (Gm(2m))/1 hat j}

color{pink} {F_(GB) =(Gm(2m))/1 (- hat i cos 30^o - hat j sin 30^o)}

color{purple} {F_(GC)= (Gm(2m))/1 (+ hat i cos 30^o - hat j sin 30^o)}

From the principle of superposition and the law of vector addition, the resultant gravitational force F_R on (2m) is

color{orange} {F_R = F_(GA)+ F_(GB) + F_(GC)}

F_(R)= 2 Gm^2 hat j + 2 Gm^2(- hat i cos 30^o - hat j sin 30^o)+ 2Gm^2(hat i cos 30^o- hat j sin 30^o)=0

Alternatively, one expects on the basis of symmetry that the resultant force ought to be zero.

(b) By symmetry the x-component of the force cancels out. The y-component survives.

F_R = 4 Gm^2 hat j- 2 Gm^2 hat j = 2Gm^2 hat j