Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर १
Lesser by a factor of 0.63
Time taken by the Earth to complete one revolution around the Sun,
Te = 1 year
Orbital radius of the Earth in its orbit, Re = 1 AU
Time taken by the planet to complete one revolution around the Sun, `T_P = 1/2T_e = 1/2` year
Orbital radius of the planet = Rp
From Kepler’s third law of planetary motion, we can write:
`(R_p/R_e)^3 = (T_p/T_e)^2`
`(R_p/R_e) = (T_p/T_e)^(2/3)`
`=((1/2)/1)^(2/3) = (0.5)^(2/3) = 0.63`
Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
उत्तर २
Here, `T_e = 1 "year"`, `T_p = T_e/2 = 1/2` year
`r_c = 1 A.U.`
Using Kepler's third law,we have `r_p = r_c (T_p/T_e)^(2/3)`
`=>r_p = 1("1/2"/1)^(2/3)` = 0.63 AU
संबंधित प्रश्न
Consider earth satellites in circular orbits. A geostationary satellite must be at a height of about 36000 km from the earth's surface. Will any satellite moving at this height be a geostationary satellite? Will any satellite moving at this height have a time period of 24 hours?
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?
At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?
Choose the correct option.
The binding energy of a satellite revolving around the planet in a circular orbit is 3 × 109 J. It's kinetic energy is ______.
Derive an expression for the binding energy of a body at rest on the Earth’s surface of a satellite.
Describe how an artificial satellite using a two-stage rocket is launched in an orbit around the Earth.
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)
The kinetic energy of a revolving satellite (mass m) at a height equal to thrice the radius of the earth (R) is ______.
There is no atmosphere on moon because ____________.
If a body weighing 40 kg-wt is taken inside the earth to a depth to `1/2` th radius of the earth, then the weight of the body at that point is ____________.
A satellite of mass 'm', revolving round the earth of radius 'r' has kinetic energy (E). Its angular momentum is ______.
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 ______.
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. If the satellite is stopped in its orbit and allowed to fall freely onto the earth, the speed with which it hits the surface ______ km/s.
[g = 9.8 ms-2 and Re = 6400 km]
The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height h above the earth's surface is ______ (R = radius of the earth).
A satellite is revolving around a planet in a circular orbit close to its surface and ρ is the mean density and R is the radius of the planet, then the period of ______.
(G = universal constant of gravitation)
Artificial satellites are launched for all the following purposes EXCEPT ______.
