Advertisements
Advertisements
प्रश्न
Calculate the critical velocity of the satellite to be located at 35780 km above the surface of earth.
उत्तर
Given: Height of the satellite above the earth’s surface (h)
= 35780 km
= 35780 × 103 m
We know that:
Gravitational constant (G) = 6.67 × 10–11 N m2/kg2,
mass of earth (M) = 6 × 1024 kg,
radius of earth (R) = 6400 km
= 6400 × 103 m
To find: Tangential velocity of satellite (vc)
Formula: `"v"_"c" = sqrt("GM"/("R + h"))`
Calculation: From formula,
`"v"_"c" = sqrt(((6.67 xx 10^-11) xx (6 xx 10^24))/((6400 + 35780) xx 10^3))`
`= sqrt((40.02 xx 10^13)/(42180 xx 10^3))`
`= sqrt(40.02/42180 xx 10^10)`
`= sqrt (0.0009487909 xx 10^10)`
`= sqrt(9.487909 xx10^6)`
≈ `sqrt9.5 xx 10^3`
= 3.08 × 103 m/s
= 3.08 km/s
The critical velocity of the satellite is 3.08 km/s.
APPEARS IN
संबंधित प्रश्न
Answer the question:
What is meant by the orbit of a satellite? On what basis and how are the orbits of artificial satellites classified?
Complete the following table.
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?
Write the proper name of the orbits of satellites shown in the following figure with their height from the earth’s surface.
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.
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 |
Define orbital velocity.
Write a note on orbital velocity
Why are some satellites called geostationary?
Numerical problem.
At an orbital height of 400 km, find the orbital period of the satellite.
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.
Write the name of small satellite made by a group of students from COEP (College of Engineering, Pune) sent to the space through ISRO in 2016.
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.)
