Advertisements
Advertisements
प्रश्न
If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking \[2\sqrt{2} T\] seconds for one revolution?
Advertisements
उत्तर
Let us assume the height of the satellite completing one revolution in 2`sqrt2` T seconds as h2.
T = `(2pi"r")/"v"_"c"`,i.e., T = `(2pi("R"+"h"_1))/(sqrt("GM"/(("R"+ "h"_1)))`
∴ T = `2pisqrt(("R"+"h"_1)^3/"GM")` ...(1)
and 2`sqrt2` T = `2pisqrt(("R"+"h"_1)^3/"GM")`
from Eqs. (1) and (2),
∴ `"T"/(2sqrt2 "T")=(2pisqrt(("R"+"h"_1)^3/"GM"))/(sqrt((("R"+ "h"_2)^3)/"GM")`
∴ `1/sqrt8=sqrt(("R"+"h"_1)^3)/(sqrt(("R"+ "h"_2)^3)`
∴ `1/2=(("R"+"h"_1))/(("R"+"h"_2)`
∴ R + h2 = 2R + 2h1
∴ h2 = R + 2h1
APPEARS IN
संबंधित प्रश्न
Solve the problem.
How much time a satellite in an orbit at height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?
Name the first artificial satellite sent by Russia in space.
The function of a satellite launcher is based on Newton's second law of motion.
Distinguish between:
High Earth orbit - Medium Earth orbit.
Note the relationship between the entries in all the three columns in the table and rewrite the table.
| Column-1 (Location) |
Column-2 Height from the earth’s surface (km) |
Column-3 g (m/s2) |
| Earth’s surface (average) | 8.8 | 0.225 |
| Mount Everest | 36.6 | 9.81 |
| The highest height ever reached by man made balloon | 400 | 9.8 |
| Orbit of spacecraft | 35700 | 9.77 |
| Orbit of communication satellite | 0 | 8.7 |
Define orbital velocity.
Write a note on orbital velocity
How is a satellite maintained in nearly circular orbit?
Numerical problem.
Calculate the speed with which a satellite moves if it is at a height of 36,000 km from the Earth’s surface and has an orbital period of 24 hr (Take R = 6370 km) [Hint: Convert hr into seconds before doing calculation]
Numerical problem.
At an orbital height of 400 km, find the orbital period of the satellite.
Write down the formula of orbital velocity.
Give scientific reasons.
The geostationary satellite is not useful in the study of polar regions.
Geostationary satellites are not useful for studies of the polar region.
Calculate the time taken by a satellite for one revolution revolving at a height of 6400 km above the earth's surface with a velocity of 5.6 km/s.
The height of medium earth orbit above the surface of the earth is ______.
The orbit of a satellite is exactly 35780 km above the earth's surface and its tangential velocity is 3.08 km/s.
How much time the satellite will take to complete one revolution around the earth?
(Radius of earth = 6400 km.)
