(4) When the distance between the two atoms becomes \[{{r}_{0}},\] the interatomic force will be zero. This distance \[{{r}_{0}}\] is called normal or equilibrium distance. \[({{r}_{0}}\text{= 0}\text{.74 { }\!\!\mathrm{\AA}\!\!\text{ } for hydrogen})\].
(5) When the distance between the two atoms further decreased, the interatomic force becomes repulsive in nature and increases very rapidly with decrease in distance between two atoms.
(6) The potential energy \[U\] is related with the interatomic force \[F\] by the following relation.
\[F=\frac{-dU}{dr}\]
(i) When two atoms are at very large distance, the potential energy is negative and becomes more negative as \[r\] is decreased.
(ii) When the distance between the two atoms becomes \[{{r}_{0,}}\] the potential energy of the system of two atoms becomes minimum (i.e. attains maximum negative value). As the state of minimum potential energy is the state of equilibrium, hence the two atoms at separation \[{{r}_{0}}\] will be in a state of equilibrium.
(\[{{U}_{0}}=-7.2\times {{10}^{-19}}\,Joule\] for hydrogen).
(iii) When the distance between the two atoms is further decreased \[(i.e.\,\,r<{{r}_{0}})\] the negative value of potential energy of the system starts decreasing. It becomes zero and then attains positive value with further decrease in \[r\] (as shown in the figure). 
(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(ii) As everything is in free fall, so objects are at rest relative to each other, i.e., if a table is withdrawn from below an object, the object will remain where it was without any support.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. | Quantities | Variation | Relation with r |
| Orbital velocity | Decreases | \[v\propto \frac{1}{\sqrt{r}}\] |
| Time period | Increases | \[T\propto {{r}^{3/2}}\] |
| Linear momentum | Decreases | \[P\propto \frac{1}{\sqrt{r}}\] |
| Angular momentum | Increases | \[L\propto \sqrt{r}\] |
| Kinetic energy | Decreases | \[K\propto \frac{1}{r}\] |
| Potential energy | Increases | more...
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.) \[=-E=\frac{GMm}{2r}\]
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.
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 \[v=3.08\ km/\sec \]
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.
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.
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
\[v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{R+h}}\] \ \[v=\sqrt{\frac{GM}{R}}=\sqrt{gR}\]
[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 \[v\propto \frac{1}{\sqrt{{{r}^{n-1}}}}\]. Current Affairs CategoriesArchive
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