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प्रश्न
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 |
| Maximum height ever reached by manmade balloon | 400 | 9.8 |
| Orbit of a typical weather satellite | 35700 | 9.77 |
| Orbit of communication satellite | 0 | 8.7 |
उत्तर
| Column-1 (Location) |
Column-2 Height from the earth’s surface (km) |
Column-3 g (m/s2) |
| Earth’s surface (average) | 0 | 9.81 |
| Mount Everest | 8.8 | 9.8 |
| Maximum height ever reached by manmade balloon | 36.6 | 9.77 |
| Orbit of a typical weather satellite | 400 | 8.7 |
| Orbit of communication satellite | 35700 | 0.225 |
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संबंधित प्रश्न
Complete the following table.
Considering first correlation, complete the second.
Hubble telescope : At 569 km above the earth’s surface
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Name the first artificial satellite sent by Russia in space.
Distinguish between:
High Earth orbit - Medium Earth orbit.
Observe the figure and write the answers.
- Name the outer orbit.
- Which satellites revolve in low earth orbits?
- Which various orbits are given in the figure?
- Give an example of a launch vehicle based on Newton’s third law.
Write functions of Military satellite and Navigational satellite.
Define 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]
Write down the formula of orbital velocity.
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.)
