Kepler’s Laws of Planetary Motion
Planets are large natural bodies rotating around a star in definite orbits. The planetary system of the star sun called solar system consists of nine planets, viz., Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto. Out of these planets Mercury is the smallest and closest to the sun and so hottest. Jupiter is largest and has maximum moons (12). Venus is closest to Earth and brightest. Kepler after a life time study, work out three empirical laws which govern the motion of these planets and are known as Kepler’s laws of planetary motion. These are,
(1) The law of Orbits : Every planet moves around the sun in an elliptical orbit with sun at one of the foci.
(2) The law of Area : The line joining the sun to the planet sweeps out equal areas in equal interval of time. i.e. areal velocity is constant. According to this law,
planet will move slowly when it is farthest from sun and more rapidly when it is nearest to sun. It is similar to law of conservation of angular momentum
Areal velocity \(=\frac{dA}{dt}=\frac{1}{2}\,\,\frac{r(vdt)}{dt}=\frac{1}{2}rv\)
\(\therefore \frac{dA}{dt}=\frac{L}{2\,m}\) [As \(L=mvr; rv=\frac{L}{m}\)]
(3) The law of periods : The square of period of revolution \((T)\) of any planet around sun is directly proportional to the cube of the semi-major axis of the orbit.
\({{T}^{2}}\,\propto \,{{a}^{3}}\) or \({{T}^{2}}\propto {{\left( \frac{{{r}_{1}}+{{r}_{2}}}{2} \right)}^{3}}\)
Proof : From the figure \(AB=AF+FB\)
\(2a={{r}_{1}}+{{r}_{2}} \,\,\, \therefore a=\frac{{{r}_{1}}+{{r}_{2}}}{2}\)
where a = semi-major axis
\({{r}_{1}}=\) Shortest distance of planet from sun (perigee).
\({{r}_{2}}=\) Largest distance of planet from sun (apogee).
Important data
Note : q Kepler's laws are valid for satellites also.
Velocity of a Planet in Terms of Eccentricity
Applying the law of conservation of angular momentum at perigee and apogee
\(m{{v}_{p}}{{r}_{p}}=m{{v}_{a}}{{r}_{a}}\)
\(\Rightarrow \frac{{{v}_{p}}}{{{v}_{a}}}=\frac{{{r}_{a}}}{{{r}_{p}}}=\frac{a+c}{a-c}=\frac{1+e}{1-e}\) [As \({{r}_{p}}=a-c,\,\,\,\,\,\,\,{{r}_{a}}=a+c\) and eccentricity \(e=\frac{c}{a}\)]
Applying the conservation of mechanical energy at perigee and apogee
\(\frac{1}{2}m{{v}_{p}}^{2}-\frac{GMm}{{{r}_{p}}}=\frac{1}{2}m{{v}_{a}}^{2}-\frac{GMm}{{{r}_{a}}}\)
\(\Rightarrow{{v}_{p}}^{2}-{{v}_{a}}^{2}=2GM\,\left[ \frac{1}{{{r}_{p}}}-\frac{1}{{{r}_{a}}} \right]\)
\(\Rightarrow{{v}_{a}}^{2}\,\left[ \frac{{{r}_{a}}^{2}-{{r}_{p}}^{2}}{{{r}_{p}}^{2}} \right]\,=\,\,2\,GM\,\left[ \frac{{{r}_{a}}-{{r}_{p}}}{{{r}_{a}}{{r}_{p}}} \right]\) [As \({{v}_{p}}=\frac{{{v}_{a}}{{r}_{a}}}{{{r}_{p}}}\)]
\(\Rightarrow{{v}_{a}}^{2}=\frac{2\,GM}{{{r}_{a}}+{{r}_{p}}}\,\,\left[ \frac{{{r}_{p}}}{{{r}_{a}}} \right]\Rightarrow{{v}_{a}}^{2}=\frac{2\,GM}{a}\,\left( \frac{a-c}{a+c} \right)\,=\,\frac{GM}{a}\,\left( \frac{1-e}{1+e} \right)\)
Thus the speeds of planet at apogee and perigee are
\({{v}_{a}}=\sqrt{\frac{GM}{a}\left( \frac{1-e}{1+e} \right)}\),
\({{v}_{p}}=\sqrt{\frac{GM}{a}\,\left( \frac{1+e}{1-e} \right)}\)
Note: The gravitational force is a central force so torque on planet relative to sun is always zero, hence angular momentum of a planet or satellite is always constant irrespective of shape of orbit.
Some Properties of the Planet
Orbital Velocity of Satellite
Satellites are natural or artificial bodies describing orbit around a planet under its gravitational attraction. Moon is a natural satellite while INSAT-1B is an artificial satellite of earth. Condition for establishment of artificial satellite is that the centre of orbit of satellite must coincide with centre of earth or satellite must move around great circle of earth.
Orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the earth.
For revolution of satellite around the earth, the gravitational pull provides the required centripetal force.
\(\frac{m{{v}^{2}}}{r}=\frac{GMm}{{{r}^{2}}}\)
\(\Rightarrow v=\sqrt{\frac{GM}{r}}\)
\(v=\sqrt{\frac{g{{R}^{2}}}{R+h}}=R\sqrt{\frac{g}{R+h}}\) [As \(GM=g{{R}^{2}}\,\, and \,\,r=R+h\)]
(i) Orbital velocity is independent of the mass of the orbiting body and is always along the tangent of the orbit i.e., satellites of diferent masses have same orbital velocity, if they are in the same orbit.
(ii) Orbital velocity depends on the mass of central body and radius of orbit.
(iii) For a given planet, greater the radius of orbit, lesser will be the orbital velocity of the satellite \(\left( v\propto 1/\sqrt{r} \right)\).
(iv) Orbital velocity of the satellite when it revolves very close to the surface of the planet
\(v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{R+h}}\) \(\therefore v=\sqrt{\frac{GM}{R}}=\sqrt{gR} \) [As \(h=0\) and \(GM=g{{R}^{2}}\)]
For the earth \(v=\sqrt{9.8\times 6.4\times {{10}^{6}}}=7.9\,k\,m/s\approx 8\,km/\sec\)
(v) Close to the surface of planet \(v=\sqrt{\frac{GM}{R}}\) [As \({{v}_{e}}=\sqrt{\frac{2GM}{R}}\)]
\(\therefore v=\frac{{{v}_{e}}}{\sqrt{2}}\) i.e.,\({{v}_{escape}}=\sqrt{2}\,\,\,{{v}_{orbital}}\)
It means that if the speed of a satellite orbiting close to the earth is made \(\sqrt{2}\) times (or increased by 41%) then it will escape from the gravitational field.
(vi) If the gravitational force of attraction of the sun on the planet varies as \(F\propto \frac{1}{{{r}^{n}}}\) then the orbital velocity varies as \(v\propto \frac{1}{\sqrt{{{r}^{n-1}}}}\).
Time Period of Satellite
It is the time taken by satellite to go once around the earth.
\(\therefore T=\frac{\text{Circumference}\ \text{of}\ \text{the}\ \text{orbit}}{\text{orbital}\ \text{velocity}}\)
\(\Rightarrow T=\frac{2\pi r}{v}=2\pi r\sqrt{\frac{r}{GM}}\) [As \(v=\sqrt{\frac{GM}{r}}\)]
\(\Rightarrow T=2\pi \,\sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \,\,\sqrt{\frac{{{r}^{3}}}{g{{R}^{2}}}}\,\,\,\,\,\,\ [As \,\,GM=g{{R}^{2}}\)]
\(\Rightarrow T=2\pi \,\,\sqrt{\frac{{{\left( R+h \right)}^{3}}}{g\,{{R}^{2}}}}=2\,\pi \,\,\sqrt{\frac{R}{g}}{{\left( 1+\frac{h}{R} \right)}^{3/2}}\) [As \(r=R+h\)]
(i) From \(T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}\), it is clear that time period is independent of the mass of orbiting body and depends on the mass of central body and radius of the orbit
(ii)\(T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}\)
\(\Rightarrow {{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{r}^{3}}\) i.e., \({{T}^{2}}\propto {{r}^{3}}\)
This is in accordance with Kepler’s third law of planetary motion \(r\) becomes a (semi major axis) if the orbit is elliptic.
(iii) Time period of nearby satellite,
From \(T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{g{{R}^{2}}}}=2\pi \sqrt{\frac{R}{g}}\) [As \(h=0\) and \(GM=g{{R}^{2}}\)]
For earth \(R=6400km\,\, and \,\,g=9.8m/{{s}^{2}}\)
\(T=84.6\,\text{minute}\ \approx 1.4\ hr\)
(iv) Time period of nearby satellite in terms of density of planet can be given as
\(T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{GM}}=\frac{2\pi {{\left( {{R}^{3}} \right)}^{1/2}}}{{{\left[ G.\frac{4}{3}\pi {{R}^{3}}\rho \right]}^{1/2}}}=\sqrt{\frac{3\pi }{G\rho }}\)
(v) If the gravitational force of attraction of the sun on the planet varies as \(F\propto \frac{1}{{{r}^{n}}}\) then the time period varies as \(T\propto {{r}^{\frac{n+1}{2}}}\)
(vi) If there is a satellite in the equatorial plane rotating in the direction of earth’s rotation from west to east, then for an observer, on the earth, angular velocity of satellite will be \(({{\omega }_{S}}-{{\omega }_{E}})\). The time interval between the two consecutive appearances overhead will be
\(T=\frac{2\pi }{{{\omega }_{s}}-{{\omega }_{E}}}=\frac{{{T}_{S}}{{T}_{E}}}{{{T}_{E}}-{{T}_{S}}}\,\,\,\,\left[ \text{As}\,\,\,T=\frac{2\pi }{\omega } \right]\)
If \({{\omega }_{S}}={{\omega }_{E}}\), \(T=\,\infty\) i.e. satellite will appear stationary relative to earth. Such satellites are called geostationary satellites.
Height of Satellite
As we know, time period of satellite \(T=2\pi \,\sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \,\,\sqrt{\frac{{{(R+h)}^{3}}}{g{{R}^{2}}}}\)
By squaring and rearranging both sides \(\frac{g\,{{R}^{2}}{{T}^{2}}}{4{{\pi }^{2}}}={{\left( R+h \right)}^{3}}\)
\(\Rightarrow h={{\left( \frac{{{T}^{2}}g\,{{R}^{2}}}{4{{\pi }^{2}}} \right)}^{1/3}}-R\)
By knowing the value of time period we can calculate the height of satellite from the surface of the earth.
Geostationary Satellite
The satellite which appears stationary relative to earth is called geostationary or geosynchronous satellite, communication satellite.
A geostationary satellite always stays over the same place above the earth such a satellite is never at rest. Such a satellite appears stationary due to its zero relative velocity w.r.t. that place on earth.
The orbit of a geostationary satellite is known as the parking orbit.
(i) It should revolve in an orbit concentric and coplanar with the equatorial plane.
(ii) Its sense of rotation should be same as that of earth about its own axis i.e., in anti-clockwise direction (from west to east).
(iii) Its period of revolution around the earth should be same as that of earth about its own axis.
\(\therefore T=24\ hr=86400\ \sec\)
(iv) Height of geostationary satellite
As \(T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}\Rightarrow 2\pi \sqrt{\frac{{{(R+h)}^{3}}}{GM}}=24hr\)
Substituting the value of \(G\) and \(M\) we get \(R+h=r=42000\ km=7R\)
\(\therefore \) height of geostationary satellite from the surface of earh\(h=6R=36000\,km\)
(v) Orbital velocity of geo stationary satellite can be calculated by \(v=\sqrt{\frac{GM}{r}}\)
Substituting the value of \(G\) and \(M\) we get \(v=3.08\ km/\sec \)
Angular Momentum of Satellite
Angular momentum of satellite \(L=mvr\)
\(\Rightarrow L=m\sqrt{\frac{GM}{r}}\ r\,\,\,\, [As \,\,v=\sqrt{\frac{GM}{r}}\)]
\(\therefore L=\sqrt{{{m}^{2}}GMr}\)
i.e., Angular momentum of satellite depends on both the mass of orbiting and central body as well as the radius of orbit.
(i) In case of satellite motion, force is central so torque = 0 and hence angular momentum of satellite is conserved i.e., \(L=\)constant
(ii) In case of satellite motion as areal velocity
\(\frac{dA}{dt}=\frac{1}{2}\,\,\frac{(r)(vdt)}{dt}=\frac{1}{2}rv\)
\(\Rightarrow\frac{dA}{dt}=\frac{L}{2\,m}\) [As \(L=mvr\)]
But as \(L=\) constant, \ areal velocity (dA/dt) = constant which is Kepler’s II law
i.e., Kepler’s II law or constancy of areal velocity is a consequence of conservation of angular momentum
Energy of Satellite
When a satellite revolves around a planet in its orbit, it possesses both potential energy (due to its position against gravitational pull of earth) and kinetic energy (due to orbital motion).
(1) Potential energy : \(U=mV=\frac{-GMm}{r}=\frac{-{{L}^{2}}}{m{{r}^{2}}}\) \(\left[ \text{As}\,\,\,V=\frac{-GM}{r},{{L}^{2}}={{m}^{2}}GMr \right]\,\)
(2) Kinetic energy : \(K=\frac{1}{2}m{{v}^{2}}=\frac{GMm}{2\,r}=\frac{{{L}^{2}}}{2\,m{{r}^{2}}}\) \(\left[ \text{As}\,\,v=\sqrt{\frac{GM}{r}} \right]\)
(3) Total energy : \(E=U+K=\frac{-GMm}{r}+\frac{GMm}{2r}=\frac{-GMm}{2r}=\frac{-{{L}^{2}}}{2m{{r}^{2}}}\)
(i) Kinetic energy, potential energy or total energy of a satellite depends on the mass of the satellite and the central body and also on the radius of the orbit.
(ii) From the above expressions we can say that
Kinetic energy (K) = – (Total energy)
Potential energy (U) = 2 (Total energy)
Potential energy (K) = – 2 (Kinetic energy)
(iii) Energy graph for a satellite
(iv) Energy distribution in elliptical orbit
(v) If the orbit of a satellite is elliptic then
(a) Total energy \((E)=\frac{-GMm}{2a}=\) constant ; where a is semi-major axis .
(b) Kinetic energy \((K)\) will be maximum when the satellite is closest to the central body (at perigee) and minimum when it is farthest from the central body (at apogee)
(c) Potential energy \((U)\) will be minimum when kinetic energy = maximum i.e., the satellite is closest to the central body (at perigee) and maximum when kinetic energy = minimum i.e., the satellite is farthest from the central body (at apogee).
(vi) Binding Energy : Total energy of a satellite in its orbit is negative. Negative energy means that the satellite is bound to the central body by an attractive force and energy must be supplied to remove it from the orbit to infinity. The energy required to remove the satellite from its orbit to infinity is called Binding Energy of the system, i.e.,
Binding Energy (B.E.) \(=-E=\frac{GMm}{2r}\)
Change in the Orbit of Satellite
When the satellite is transferred to a higher orbit \(({{r}_{2}}\) >\({{r}_{1}}\) ) then variation in different quantities can be shown by the following table
Note: Work done in changing the orbit \(W={{E}_{2}}-{{E}_{1}}\)
\(W=\left( -\frac{GMm}{2{{r}_{2}}} \right)-\left( -\frac{GMm}{2{{r}_{1}}} \right)\)
\(W=\frac{GMm}{2}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\)
Weightlessness
The weight of a body is the force with which it is attracted towards the centre of earth. When a body is stationary with respect to the earth, its weight equals the gravity. This weight of the body is known as its static or true weight.
We become conscious of our weight, only when our weight (which is gravity) is opposed by some other object. Actually, the secret of measuring the weight of a body with a weighing machine lies in the fact that as we place the body on the machine, the weighing machine opposes the weight of the body. The reaction of the weighing machine to the body gives the measure of the weight of the body.
The state of weightlessness can be observed in the following situations.
(1) When objects fall freely under gravity : For example, a lift falling freely, or an airship showing a feat in which it falls freely for a few seconds during its flight, are in state of weightlessness.
(2) When a satellite revolves in its orbit around the earth : Weightlessness poses many serious problems to the astronauts. It becomes quite difficult for them to control their movements. Everything in the satellite has to be kept tied down. Creation of artificial gravity is the answer to this problem.
(3) When bodies are at null points in outer space : On a body projected up, the pull of the earth goes on decreasing, but at the same time the gravitational pull of the moon on the body goes on increasing. At one particular position, the two gravitational pulls may be equal and opposite and the net pull on the body becomes zero. This is zero gravity region or the null point and the body in question is said to appear weightless.
Weightlessness in a Satellite
A satellite, which does not produce its own gravity moves around the earth in a circular orbit under the action of gravity. The acceleration of satellite is \(\frac{GM}{{{r}^{2}}}\) towards the centre of earth.
If a body of mass m placed on a surface inside a satellite moving around the earth. Then force on the body are
(i) The gravitational pull of earth \(=\frac{GMm}{{{r}^{2}}}\)
(ii) The reaction by the surface \(=R\)
By Newton’s law \(\frac{GmM}{{{r}^{2}}}-R=m\ a\)
\(\frac{GmM}{{{r}^{2}}}-R=m\ \left( \frac{GM}{{{r}^{2}}} \right)\)
\(\therefore\) R = 0
Thus the surface does not exert any force on the body and hence its apparent weight is zero.
A body needs no support to stay at rest in the satellite and hence all position are equally comfortable. Such a state is called weightlessness.
(i) One will find it difficult to control his movement, without weight he will tend to float freely. To get from one spot to the other he will have to push himself away from the walls or some other fixed objects.
(ii) As everything is in free fall, so objects are at rest relative to each other, i.e., if a table is withdrawn from below an object, the object will remain where it was without any support.
(iii) If a glass of water is tilted and glass is pulled out, the liquid in the shape of container will float and will not flow because of surface tension.
(iv) If one tries to strike a match, the head will light but the stick will not burn. This is because in this situation convection currents will not be set up which supply oxygen for combustion
(v) If one tries to perform simple pendulum experiment, the pendulum will not oscillate. It is because there will not be any restoring torque and so \(T=2\pi \sqrt{(L/{g}')}=\infty\). [As \({g}'=0\)]
(vi) Condition of weightlessness can be experienced only when the mass of satellite is negligible so that it does not produce its own gravity.
e.g. Moon is a satellite of earth but due to its own weight it applies gravitational force of attraction on the body placed on its surface and hence weight of the body will not be equal to zero at the surface of the moon.