Advertisements
Advertisements
Question
Draw a labelled diagram to show different trajectories of a satellite depending upon the tangential projection speed.
Advertisements
Solution
vh = horizontal speed of projection
vc = critical velocity
ve = escape velocity
APPEARS IN
RELATED QUESTIONS
Suppose there existed a planet that went around the sun twice as fast as the earth.What would be its orbital size as compared to that of the earth?
Is it necessary for the plane of the orbit of a satellite to pass through the centre of the earth?
Two satellites going in equatorial plane have almost same radii. As seen from the earth one moves from east one to west and the other from west to east. Will they have the same time period as seen from the earth? If not which one will have less time period?
The time period of an earth-satellite in circular orbit is independent of
A body stretches a spring by a particular length at the earth's surface at the equator. At what height above the south pole will it stretch the same spring by the same length? Assume the earth to be spherical.
A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth's surface. Find (a) its speed in the orbit, (b) is kinetic energy, (c) the potential energy of the earth-satellite system and (d) its time period. Mass of the earth = 6 × 1024kg.
Answer the following question.
Define the binding energy of a satellite.
Answer the following question.
What is periodic time of a geostationary satellite?
State the conditions for various possible orbits of satellite depending upon the horizontal/tangential speed of projection.
Answer the following question in detail.
What is a critical velocity?
Answer the following question in detail.
Two satellites A and B are revolving round a planet. Their periods of revolution are 1 hour and 8 hour respectively. The radius of orbit of satellite B is 4 × 104 km. Find radius of orbit of satellite A.
Solve the following problem.
Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface.
(ME = 5.98 × 1024 kg, R = 6.4 × 106 m)
Solve the following problem.
Calculate the value of acceleration due to gravity on the surface of Mars if the radius of Mars = 3.4 × 103 km and its mass is 6.4 × 1023 kg.
A body weighs 5.6 kg wt on the surface of the Earth. How much will be its weight on a planet whose mass is 7 times the mass of the Earth and radius twice that of the Earth’s radius?
The ratio of energy required to raise a satellite of mass 'm' to a height 'h' above the earth's surface of that required to put it into the orbit at same height is ______.
[R = radius of the earth]
What is the minimum energy required to launch a satellite of mass 'm' from the surface of the earth of mass 'M' and radius 'R' at an altitude 2R?
If the Earth-Sun distance is held constant and the mass of the Sun is doubled, then the period of revolution of the earth around the Sun will change to ____________.
Out of following, the only correct statement about satellites is ____________.
A satellite of mass 'm' is revolving around the earth of mass 'M' in an orbit of radius 'r' with constant angular velocity 'ω'. The angular momentum of the satellite is ______.
(G =gravitational constant)
In the case of earth, mean radius is 'R', acceleration due to gravity on the surface is 'g', angular speed about its own axis is 'ω'. What will be the radius of the orbit of a geostationary satellite?
A satellite is to revolve round the earth in a circle of radius 9600 km. The speed with which this satellite be projected into an orbit, will be ______.
Two satellites are orbiting around the earth in circular orbits of same radius. One of them is 10 times greater in mass than the other. Their period of revolutions are in the ratio ______.
Which of the following is an example of a communication (geostationary) satellite launched by India?
Which application is mainly associated with polar satellites?
A satellite is revolving round the earth with orbital speed ‘V0’. If it stops suddenly, the speed with which it will strike the surface of the earth would be: (V = escape velocity of a particle on earth’s surface)
