Gravitation | Conceptual notes | Important Questions | Short Questions and Numerical problem Solutions | Class 11 (Physics) | Physics in Depth
Newton's universal law of gravitation
The force of attraction between two given particles is inversely proportional to the square of their distance apart.Dimension of G = [L3M-1T-2] and \(G=6.67\times 10^{-11} Nm^2 kg^{-2}\)
Newton's law of gravitation is also called the universal law as it is applicable for both celestial and terrestrial bodies in the universe.
Inertial mass
It is the measure of ability of body to oppose the change in the velocity(i.e., acceleration produced by an external force). From Newton's second law of motion,Thus, inertial mass of the body is equal to the magnitude of force to produce unit acceleration in a body.
Gravitational mass
It is defined by the Newton's universal law of gravitation.No experiment has unambiguously detected any difference between gravitational and inertial mass. In classical physics , it's not clear why these two are same.
Variation of acceleration of free fall
For points outside the earth, the gravitational force obeys an inverse-square law. So the acceleration of free fall, \(g' \propto \frac{1}{r^2}\), where r is the distance to the center of the earth from the body. The maximum value of \(\color{red}g'\) is obtained at the earth's surface where \(\color{red}r=R\) ; R is the radius of earth. Inside the earth, the value of \(\color{red}g'\) is not inversely proportional to the square of the distance from the earth. Assuming a uniform earth density (not true in practice though), we now derive the expression for \(\color{red}g'\) inside the earth.
on earth's surface, \(g=\frac{GM}{R^2}=\frac{G\frac{4}{3}\pi R^3 \rho}{R^2}\) \(\therefore g=\frac{4}{3} G\pi R \rho\)
Let us consider a point at a depth x from the earth's surface, then acceleration due to gravity at that point due to the sphere of radius (R-x).
Weightlessness
Suppose R is the reaction of the couch to which the astronaut is initially strapped. Then from \(F=ma\), \(R-mg=ma=m.15g\), where m is the mass of the astronaut. Then R=16mg. This force is 16 times the weight of the astronaut and thus, initially, he experiences a large force.
In orbit, however, the state of affairs is different. This time the acceleration of the spacecraft and astronaut are both \({\color{red}g'}\) in magnitude, where \({\color{red}g'}\) is the acceleration due to gravity at the particular height of the orbit. If R' is the reaction of the surface of the spacecraft in contact with the astronaut , then, for circular motion,
At earth's surface, we feel the reaction at the ground and are thus conscious of our weight. Inside a freely falling lift, the reaction at our feet diminishes and we feel weightless.
Gravitational field and Gravitational field intensity
Gravitational field
The area round about a body within which its force of gravitational attraction is perceptible (no other body being near about it) is called its gravitational field.Gravitational field intensity
The gravitational field intensity, E, of the gravitational field of a particle of mass M at a point distant r from it is the force experienced by a unit mass placed at that point in the field.Gravitational potential and Gravitational potential energy
Gravitational potential
The gravitational potential V at a point distant r from a body of mass M is equal to the amount of work done in moving a unit mass from infinity (where the gravitational force and potential energy are zero ) to that point.Gravitational potential energy
The gravitational potential energy U at a point distant r from a body of mass M is the amount of work done in moving an object of mass m from infinity to that point.Satellites
The orbital velocity of the satellite at height h from the earth's surface is \[\color{green} v_0=R\sqrt{\frac{g}{R+h}}\] where, R is the radius of earth and if the satellite is very close to earth's surface, \(\color{red} v_0=\sqrt{gR}\).
An earth satellite moves in a circular orbit with a speed of 6.2 km/s. Find the time of one revolution and its centripetal acceleration.
Here, orbital velocity \(v_0\) = 6.2 km/s = 6200 m/s, radius of earth R = 6400 km = 6400000 m, time of one revolution T = ? and centripetal acceleration a = ?
Let, r be the distance of satellite from the center of earth.
Now, the orbital velocity is,
\[\begin{align*}
v_0&=R\sqrt{\frac{g}{r}}\\
6200&=6400000 \times \sqrt{\frac{10}{r}}\\
r&=10655567.12 \hspace{0.1cm} m
\end{align*}\]
Now, the time period is,
\[\begin{align*}
T&=\frac{2\pi r}{v_0}\\
&=\frac{2\pi \times 10655567.12}{6200}\\
&=10798.53\hspace{0.1cm} s\\
&=2.99 \hspace{0.1cm} \text{hrs}
\end{align*}\]
Then, the centripetal acceleration is,
\[a=\omega^2 r= \left(\frac{2\pi}{T}\right)^2 \times 10655567.12=3.61 \hspace{0.1cm} m/s^2\]
Parking orbits and geostationary satellite
If the period of the satellite in its orbit is exactly equal to the period of the earth as it turns about its axis, which is approx. 24 hrs, the satellite will stay over the same place on the earth while the earth rotates . This is sometimes called the parking orbit and these satellites are called geostationary satellite . Relay satellites can be placed in parking orbits, so that television programmes can be transmitted continuously from one part of the world to another.Escape velocity
The escape velocity is defined as the speed that an object must travel to break free of a planet's gravitational pull. To escape from a gravitational pull of the earth, for example, a rocket needs a speed of 11 km/s (calculated from \({v_e}=\sqrt{2gR}\)).
If an object gains enough speed to attain escape velocity, its orbit becomes an open curve called a parabola. If it continues moving faster than escape velocity, its orbit is a flattened curve called a hyperbola. A spacecraft that leaves its orbit around the Earth on a journey toward another planet travels in a hyperbolic orbit.Why is there lack of atmosphere round the moon?
The escape velocity of moon is only 2.4 km/s as against 11.2 km/s on the earth. Then, all the constituents of the atmosphere like oxygen, nitrogen, carbondioxide and water vapour, the root mean square velocities of which at 00 C lie between 0.4 km/s and 0.8 km/s - also the hydrogen and helium, whose r.m.s. velocities are very much greater , about 2 km/s should all be absent from the surface of the moon. Thus, there is lack of atmosphere in moon. NOte that!a planet can hold onto it's atmosphere if the escape velocity is about 4-6 times greater than the average kinetic velocity.
Energy of a satellite
As a satellite of mass m orbits Earth, both its speed, which fixes its kinetic energy K.E, and its distance from the center of Earth, which fixes its gravitational potential energy , P.E., changes with time but the total energy, E, remains constant.Black holes
Black hole is a region of space having a gravitational field so intense that no matter or radiation can escape from it. A star with a mass greater than 20 times the mass of our Sun may produce a black hole at the end of its life.Characteristics:
- The ‘event horizon’ is the boundary defining the region of space around a black hole from which nothing (not even light) can escape i.e., the escape velocity for an object within the event horizon exceeds the speed of light.
- The Schwarzschild radius is the radius of the event horizon surrounding a non-rotating black hole. Any object with a physical radius smaller than its Schwarzschild radius will be a black hole. It is given as,\(R_s=\frac{2GM}{c^2}\)
- Black hole has three independent physical properties: mass, charge and angular momentum.
Click on Gravitation_questions to find the important questions of ' Gravitation'.
Click on Gravitation_solutions to find the solutions to short questions and numerical problems.
Comments
Post a Comment