1.The escape velocity of a body on an imaginary planet which has thrice the radius of the earth and twice the mass of the earth is (where ve is the escape velocity on the earth)
(a) √(2/3).ve
(b) √(3/2).ve
(c) √(2).ve/3
(d) 2ve/√3
2.The orbital velocity of an artificial satellite near the surface of the moon is increased by 41.4%. The satellite will
(a) move in an orbit of radius greater by 41.4%
(b) move in an orbit of radius twice the original value
(c) move in an elliptical orbit
(d) escape into outer space
3.The orbital speed of an artificial satellite very close to the surface of the earth is V0. Then the orbital speed of another artificial satellite at a height equal to 3 times the radius of the earth is
(a) 4V0
(b) 2V0
(c) V0
(d) 0.5V0
4.A simple pendulum has a time period T1 when on earth’s surface, and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T2/T1 is
(a) 1
(b) √2
(c) 4
(d) 2
5.If a body of mass ‘m’ is raised from the surface of the earth to a height ‘h’ which is comparable to the radius of the earth R, the work done is
(a) mgh
(b) mgh[1-(h/R)]
(c) mgh[1+(h/R)]
(d) mgh/[1+(h/R)]
Showing posts with label Gravitation. Show all posts
Showing posts with label Gravitation. Show all posts
Satellites
FOR MORE ON GRAVITATION CLICK HERE
Newton’s concept of a satellite's orbit
•Newton created a hypothetical scenario as follows. A person climbed a very tall mountain and launched a projectile horizontally from the peak. The projectile follows a parabolic path (see the above discussion relating to projectile motion) before striking the ground. If another projectile were launched faster than the first, then it would travel further before striking the ground. If yet another projectile were launched fast enough, then it should be able to travel right around the Earth because, as it falls, the surface of the Earth curves away from it. The curve of the projectile’s motion would follow that of the earth’s surface and thus not hit it. This projectile would then be in a circular orbit at a fixed height above the earth’s surface.
•If a projectile is launched still faster, its orbit will stretch out into an elliptical shape. Even faster launch velocities result in the projectile following a parabolic or hyperbolic path away from the Earth, escaping it entirely.
Circular Velocity (Orbital Velocity)
where r = R+h
At the earth's surface :
in this case the binding energy is equal to (GMm)/R.
For a satellite in circular orbit around the Earth:
Escape Velocity:
Newton’s concept of a satellite's orbit
•Newton created a hypothetical scenario as follows. A person climbed a very tall mountain and launched a projectile horizontally from the peak. The projectile follows a parabolic path (see the above discussion relating to projectile motion) before striking the ground. If another projectile were launched faster than the first, then it would travel further before striking the ground. If yet another projectile were launched fast enough, then it should be able to travel right around the Earth because, as it falls, the surface of the Earth curves away from it. The curve of the projectile’s motion would follow that of the earth’s surface and thus not hit it. This projectile would then be in a circular orbit at a fixed height above the earth’s surface.
•If a projectile is launched still faster, its orbit will stretch out into an elliptical shape. Even faster launch velocities result in the projectile following a parabolic or hyperbolic path away from the Earth, escaping it entirely.
Circular Velocity (Orbital Velocity)
Orbital velocity (also called circular velocity) is the speed of a satellite in orbit around a planet. The formula for circular velocity is given as:
where r = R+hTime period of Satellite
Geostationary satellite
A satellite whose period of revolution is 24 hours, is a geostationary satellite
It always appears to be at a fixed point in space, because the period of rotation of the Earth about its own axis is also equal to 24 hours.
Knowing T = 24 hours, g = 9.8 ms-1, the height of a geostationary satellite is calculated to be 36000km
Its orbital velocity is 3.1 km/s
Its plane of orbit is the equatorial plane
It revolves from west to east which is similar to the Earth's movement.
It is very useful in telecommunication.
Binding Energy
Total energy of an object within the gravitational field of the Earth is always given by the expression :
ETOTAL = EK + EP (Conservation of Energy)
At the earth's surface :
EK = 0
Therefore ETOTAL = EP = -(GMm)/R,
in this case the binding energy is equal to (GMm)/R.
For a satellite in circular orbit around the Earth:
The minimum velocity required by a satellite or a rocket or any other object to escape the Earth's Gravitational Potential Energy is called escape velocity (ve)
For Earth the value of escape velocity is 11.2 km/s
GRAVITATION: Points to Remember
- Newton's law of gravitation
- Acceleration due to gravity
- Factors effecting Acceleration due to Gravity
- Gravitational Field
- Gravitational Potential
- Kepler's Laws
- Satellites
Newton's law of gravitation
F α 1/r2
F = G(m1m2/r2)
where G = Universal Gravitational const
G = 6.67 x 10-11Nm2/kg2
Acceleration due to gravity
On the Surface
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| THE COIN AND FEATHER EXPERIMENT |
Point to note : Acceleration due to gravity is independent of the mass of the object
Factors effecting acceleration due to gravity
Altitude
With increase in altitude the value of 'g' decreases
Depth
With increase in depth the value of 'g' decreases.
 |
| When the particle is at depth 'd' mass of only the shaded part is considered |
Assuming a planet to be a perfect sphere having a uniform density 'ρ' and radius 'R'
Acceleration due to Gravity at the surface = g = (4/3)πρGR
At a depth 'd' from the surface of the planet
Acceleration due to gravity will be = gd = (4/3)πρG(R-d)
Note: 'g' at the centre of planet = 0
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| The plot shows variations in gravitational acceleration as we move away from center of Planet |
Rotation of planet(Effect of latitude)
Assume the Earth to be a sphere of radius R and mass M rotating on its axis with an angular velocity w. Consider a particle of mass m at P such that OP makes an angle of ø with OE . Here ø is the latitude of the particle. The particle is moving in a circle of radius R' = R cos ø.
The net force pulling the particle towards the center of the Earth is
where FN is the total force acting on the particle, FG is the force due to gravity and Fc is the centrifugal force. The centrifugal force is given by
If g' represents the gravitational acceleration, then
Note : At poles ø = 900 : g' = g
At equator ø = 00 : g' = g - Rω2
Gravitational field
It is assumed that a body say A, creates a gravitational field in the space around it. The field has its own existence and has energy and momentum. When another body B is placed in gravitational field of a body, this field exerts a force on it. The direction and intensity of the field is defined in terms of the force it exerts on a body placed in it.
The intensity of gravitational field vector E at a point is defined by the equation
E = F/mass
where F is the force vector exerted by the field on a body of mass m placed in the field. The intensity of gravitational field is abbreviated as gravitational field. Its SI unit is N/kg.
By the way they are defined, intensity of gravitational field and acceleration due to gravity have equal magnitudes and directions, but they are two separate physical quantities.
Gravitational potential
Gravitational potential at a point is equal to the change in potential energy per unit mass, as the mass is brought from the reference point to the given point.
Es = - (GM/r)
Es = - (GMm/r)
Gravitational Potental Energy
Es = - (GMm/r)
1. All planets move in elliptical orbits with the sun at a focus.
2. The radius vector from the sun to the planet sweeps equal area in equal time.
3. The square of the time period of a plant is proportional to the cube of the semimajor axis of the ellipse.
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