Explanation of Universal Law Of Gravitation

Explanation of Universal Law Of Gravitation

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

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaBa % aaleaacaWGTbaabeaakiabg2da9maalaaabaGaamOvamaaCaaaleqa % baGaaGOmaaaaaOqaaiaadkfadaWgaaWcbaGaamyBaaqabaaaaOGaey % ypa0ZaaSaaaeaacaaI0aGaeqiWda3aaWbaaSqabeaacaaIYaaaaOGa % amOuamaaBaaaleaacaWGTbaabeaaaOqaaiaadsfadaahaaWcbeqaai % aaikdaaaaaaaaa!452C! {a_m} = \frac{{{V^2}}}{{{R_m}}} = \frac{{4{\pi ^2}{R_m}}}{{{T^2}}}\)           (8.3)

where V is the speed of the moon related to the time period T by the relation V = 2\(\pi\) R_m / T . The time period T is about 27.3 days and R _m was already known then to be about 3.84 × 10⁸m.If we substitute these numbers in Eq. (8.3), we get a value of a_m 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 centre of the earth, we will have a_m \(\alpha\)R\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaai % aad2gaaeaacaaIYaaaaaaa!37D1! _m^{-2}\) ; g \(\alpha\) R\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaai % aadweaaeaacaaIYaaaaaaa!37A9! _E^{-2}\) and we get

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGNbaabaGaamyyamaaBaaaleaacaWGTbaabeaaaaGccqGH9aqpdaWc % aaqaaiaadkfadaqhaaWcbaGaamyBaaqaaiaaikdaaaaakeaacaWGsb % Waa0baaSqaaiaadweaaeaacaaIYaaaaaaakiabloKi7iaaiodacaaI % 2aGaaGimaiaaicdaaaa!4388! \frac{g}{{{a_m}}} = \frac{{R_m^2}}{{R_E^2}} \simeq 3600\)               (8.4)

in agreement with a value of g 9.8 m s^-2 and the value of am from Eq. (8.3). 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 m₂ due to another point mass m₁ has the magnitude 

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WGgbaacaGLhWUaayjcSdGaeyypa0Jaam4ramaalaaabaGaamyBamaa % BaaaleaacaaIXaaabeaakiaad2gadaWgaaWcbaGaaGOmaaqabaaake % aacaWGYbWaaWbaaSqabeaacaaIYaaaaaaaaaa!416C! \left| F \right| = G\frac{{{m_1}{m_2}}}{{{r^2}}}\)                                  (8.5)

Equation (8.5) can be expressed in vector form as

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGgb % Gaeyypa0Jaam4ramaalaaabaGaamyBamaaBaaaleaacaaIXaaabeaa % kiaad2gadaWgaaWcbaGaaGOmaaqabaaakeaacaWGYbWaaWbaaSqabe % aacaaIYaaaaaaakmaabmaabaGaeyOeI0IabmOCayaajaaacaGLOaGa % ayzkaaGaeyypa0JaeyOeI0Iaam4ramaalaaabaGaamyBamaaBaaale % aacaaIXaaabeaakiaad2gadaWgaaWcbaGaaGOmaaqabaaakeaacaWG % YbWaaWbaaSqabeaacaaIYaaaaaaakiqadkhagaqcaaqaaiabg2da9i % abgkHiTiaadEeadaWcaaqaaiaad2gadaWgaaWcbaGaaGymaaqabaGc % caWGTbWaaSbaaSqaaiaaikdaaeqaaaGcbaWaaqWaaeaacaWGYbaaca % GLhWUaayjcSdWaaWbaaSqabeaacaaIZaaaaaaakiqadkhagaqcaaaa % aa!5809! \begin{array}{l} F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\left( { - \hat r} \right) = - G\frac{{{m_1}{m_2}}}{{{r^2}}}\hat r\\ = - G\frac{{{m_1}{m_2}}}{{{{\left| r \right|}^3}}}\hat r \end{array}\)

where G is the universal gravitational constant,\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOCayaaja % aaaa!36FD! {\hat r}\) is the unit vector from m₁  to m₂  and r = r₂  – r₁ as shown in fig.8.3

Fig. 8.3 Gravitational force on m₁ due to m₂ is along r where the vector r is (r₂– r₁).

The gravitational force is attractive, i.e., the force F is along – r. The force on point mass m₁ due to m₂ is of course – F by Newton’s third law.

Thus, the gravitational force F₁₂ on the body 1 due to 2 and F₂₁ on the body 2 due to 1 are related as F₁₂   = – 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.

Fig. 8.4 Gravitational force on point mass m is the vector sum of the gravitational forces exerted by m₂, m₃ and m₄.The total force m₁ is

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa % aaleaacaaIXaaabeaakiabg2da9maalaaabaGaam4raiaad2gadaWg % aaWcbaGaaGOmaaqabaGccaWGTbWaaSbaaSqaaiaaigdaaeqaaaGcba % GaamOCamaaDaaaleaacaaIYaGaaGymaaqaaiaaikdaaaaaaOGabmOC % ayaajaWaaSbaaSqaaiaaikdacaaIXaaabeaakiabgUcaRmaalaaaba % Gaam4raiaad2gadaWgaaWcbaGaaG4maaqabaGccaWGTbWaaSbaaSqa % aiaaigdaaeqaaaGcbaGaamOCamaaDaaaleaacaaIZaGaaGymaaqaai % aaikdaaaaaaOGabmOCayaajaWaaSbaaSqaaiaaiodacaaIXaaabeaa % kiabgUcaRmaalaaabaGaam4raiaad2gadaWgaaWcbaGaaGinaaqaba % GccaWGTbWaaSbaaSqaaiaaigdaaeqaaaGcbaGaamOCamaaDaaaleaa % caaI0aGaaGymaaqaaiaaikdaaaaaaOGabmOCayaajaWaaSbaaSqaai % aaisdacaaIXaaabeaaaaa!5AA3! {F_1} = \frac{{G{m_2}{m_1}}}{{r_{21}^2}}{\hat r_{21}} + \frac{{G{m_3}{m_1}}}{{r_{31}^2}}{\hat r_{31}} + \frac{{G{m_4}{m_1}}}{{r_{41}^2}}{\hat r_{41}}\)

EXAMPLE 2 

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

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

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

Fig. 8.5 Three equal masses are placed at the three vertices of the \(\Delta\) ABC. A mass 2m is placed at the centroid G.

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGgb % WaaSbaaSqaaiaadEeacaWGbbaabeaakiabg2da9maalaaabaGaam4r % aiaad2gadaqadaqaaiaaikdacaWGTbaacaGLOaGaayzkaaaabaGaaG % ymaaaaceWGQbGbaKaaaeaacaWGgbWaaSbaaSqaaiaadEeacaWGbbaa % beaakiabg2da9maalaaabaGaam4raiaad2gadaqadaqaaiaaikdaca % WGTbaacaGLOaGaayzkaaaabaGaaGymaaaadaqadaqaaiabgkHiTiqa % dMgagaqcaiGacogacaGGVbGaai4CaiaaiodacaaIWaWaaWbaaSqabe % aacaaIWaaaaOGaeyOeI0IabmOAayaajaGaci4CaiaacMgacaGGUbGa % aG4maiaaicdadaahaaWcbeqaaiaaicdaaaaakiaawIcacaGLPaaaae % aacaWGgbWaaSbaaSqaaiaadEeacaWGdbaabeaakiabg2da9maalaaa % baGaam4raiaad2gadaqadaqaaiaaikdacaWGTbaacaGLOaGaayzkaa % aabaGaaGymaaaadaqadaqaaiabgUcaRiqadMgagaqcaiGacogacaGG % VbGaai4CaiaaiodacaaIWaWaaWbaaSqabeaacaaIWaaaaOGaeyOeI0 % IabmOAayaajaGaci4CaiaacMgacaGGUbGaaG4maiaaicdadaahaaWc % beqaaiaaicdaaaaakiaawIcacaGLPaaaaaaa!72B5! \begin{array}{l} {F_{GA}} = \frac{{Gm\left( {2m} \right)}}{1}\hat j\\ {F_{GA}} = \frac{{Gm\left( {2m} \right)}}{1}\left( { - \hat i\cos {{30}^0} - \hat j\sin {{30}^0}} \right)\\ {F_{GC}} = \frac{{Gm\left( {2m} \right)}}{1}\left( { + \hat i\cos {{30}^0} - \hat j\sin {{30}^0}} \right) \end{array}\)

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

F_R = F_GA + F_GB + F_GC

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaaBa % aaleaacaWGsbaabeaakiabg2da9iaaikdacaWGhbGaamyBamaaCaaa % leqabaGaaGOmaaaakiqadQgagaqcaiabgUcaRiaaikdacaWGhbGaam % yBamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaeyOeI0IabmyAayaa % jaGaci4yaiaac+gacaGGZbGaaG4maiaaicdadaahaaWcbeqaaiaaic % daaaGccqGHsislceWGQbGbaKaaciGGZbGaaiyAaiaac6gacaaIZaGa % aGimamaaCaaaleqabaGaaGimaaaaaOGaayjkaiaawMcaaiabgUcaRi % aaikdacaWGhbGaamyBamaaCaaaleqabaGaaGOmaaaakmaabmaabaGa % bmyAayaajaGaci4yaiaac+gacaGGZbGaaG4maiaaicdadaahaaWcbe % qaaiaaicdaaaGccqGHsislceWGQbGbaKaaciGGZbGaaiyAaiaac6ga % caaIZaGaaGimamaaCaaaleqabaGaaGimaaaaaOGaayjkaiaawMcaai % abg2da9iaaicdaaaa!6667! {F_R} = 2G{m^2}\hat j + 2G{m^2}\left( { - \hat i\cos {{30}^0} - \hat j\sin {{30}^0}} \right) + 2G{m^2}\left( {\hat i\cos {{30}^0} - \hat j\sin {{30}^0}} \right) = 0\)

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

(b) Now if the mass at vertex A is doubled then

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaceWGgb % GbauaadaWgaaWcbaGaam4raiaadgeaaeqaaOGaeyypa0ZaaSaaaeaa % caWGhbGaaGOmaiaad2gacaGGUaGaaGOmaiaad2gaaeaacaaIXaaaai % qadQgagaqcaiabg2da9iaaisdacaWGhbGaamyBamaaCaaaleqabaGa % aGOmaaaakiqadQgagaqcaaqaaiqadAeagaqbamaaBaaaleaacaWGhb % GaamOqaaqabaGccqGH9aqpcaWGgbWaaSbaaSqaaiaadEeacaWGcbaa % beaakiaaykW7caaMc8UaaGPaVlaadggacaWGUbGaamizaiaaykW7ca % aMc8UaaGPaVlaaykW7ceWGgbGbauaadaWgaaWcbaGaam4raiaadoea % aeqaaOGaeyypa0JaamOramaaBaaaleaacaWGhbGaam4qaaqabaaake % aaceWGgbGbauaadaWgaaWcbaGaamOuaaqabaGccqGH9aqpceWGgbGb % auaadaWgaaWcbaGaam4raiaadgeaaeqaaOGaey4kaSIabmOrayaafa % WaaSbaaSqaaiaadEeacaWGcbaabeaakiabgUcaRiqadAeagaqbamaa % BaaaleaacaWGhbGaam4qaaqabaaakeaaceWGgbGbauaadaWgaaWcba % GaamOuaaqabaGccqGH9aqpcaaIYaGaam4raiaad2gadaahaaWcbeqa % aiaaikdaaaGcceWGQbGbaKaaaaaa!73A5! \begin{array}{l} {{F'}_{GA}} = \frac{{G2m.2m}}{1}\hat j = 4G{m^2}\hat j\\ {{F'}_{GB}} = {F_{GB}}\,\,\,and\,\,\,\,{{F'}_{GC}} = {F_{GC}}\\ {{F'}_R} = {{F'}_{GA}} + {{F'}_{GB}} + {{F'}_{GC}}\\ {{F'}_R} = 2G{m^2}\hat j \end{array}\)

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 point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell.Qualitatively this can be understood as follows: 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.

(2)The force of attraction due to a hollow spherical shell of uniform density, on a 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.

Central Forces

Central Forces

We know the time rate of change of the angular momentum of a single particle about the origin is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGKbGaaGymaaqaaiaadsgacaWG0baaaiaadkhacqGHxdaTcaWGgbaa % aa!3D65! \frac{{d1}}{{dt}}r \times F\)

The angular momentum of the particle is conserved, if the torque \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiXdqNaey % 41aqRaamOCaiaadAeaaaa!3B94! \tau \times rF\) due to the force F on it vanishes. This happens either when F is zero or when F is along r. We are interested in forces which satisfy the latter condition. Central forces satisfy this condition.

A ‘central’ force is always directed towards or away from a fixed point, i.e., along the position vector of the point of application of the force with respect to the fixed point. (See Figure below.) Further, the magnitude of a central force F depends on r, the distance of the point of application of the force from the fixed point; F = F(r).

In the motion under a central force the angular momentum is always conserved. Two important results follow from this:

1. The motion of a particle under the central force is always confined to a plane.

2.The position vector of the particle with respect to the centre of the force (i.e. the fixed point) has a constant areal velocity. In other words the position vector sweeps out equal areas in equal times as the particle moves under the influence of the central force.

Try to prove both these results. You may need to know that the areal velocity is given by : dA/dt = ½ r v sin \(\alpha\).

An immediate application of the above discussion can be made to the motion of a planet under the gravitational force of the sun. For convenience the sun may be taken to be so heavy that it is at rest. The gravitational force of the sun on the planet is directed towards the sun.

This force also satisfies the requirement F = F(r), since F = G m₁ m₂ /r² where m and m are respectively the masses of the planet and the sun and G is the universal constant of gravitation.

The two results (1) and (2) described above, therefore, apply to the motion of the planet. In fact, the result (2) is the well-known second law of Kepler.

Tr is the trejectory of the particle under the central force. At a position P, the force is directed along OP, O is the centre of the force taken as the origin. In time \(\Delta\)t, the particle moves from P to \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmiuayaafa % aaaa!36D7! P'\), arc \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiqadc % fagaqbaaaa!37AC! PP'\) = \(\Delta\)s = v \(\Delta\)t. The tangent PQ at P to the trajectory gives the direction of the velocity at P. The area swept in \(\Delta\)t is the area of sector \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiaad+ % eaceWGqbGbauaacqGHijYUdaqadaqaaiaadkhaciGGZbGaaiyAaiaa % c6gacqaHXoqyaiaawIcacaGLPaaacaWGqbGabmiuayaafaGaai4lai % aaikdacqGH9aqpdaqadaqaaiaadkhacaWG2bGaci4CaiaacMgacaGG % UbGaamyyaaGaayjkaiaawMcaaiabgs5aejaadshacaGGVaGaaGOmaa % aa!505B! POP' \approx \left( {r\sin \alpha } \right)PP'/2 = \left( {rv\sin a} \right)\Delta t/2\).)