Advertisements
Advertisements
प्रश्न
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.)
Advertisements
उत्तर
Given:
Height of the satellite above the earth’s surface = 35780 km.
Tangential velocity of the satellite = 3.08 km/sec
Suppose the satellite takes T seconds to complete one revolution around the Earth. The distance travelled during this one revolution is equal to the circumference of the circular orbit. If r is the radius of the orbit, the satellite will travel a distance of 2πr during one revolution. Thus, the time required for one complete revolution can be obtained as follows:
v = `"distance"/"time"`
= `"circumference"/"time"`
= `(2pir)/"T"` ...[∵ x = (r + h)]
T = `(2pir)/v`
= `2pi("R" + h)`
= `(2 xx 3.14 xx (6400 + 35780))/3.08`
= 86003.38 sec
= 23.89 hrs.
= 23 hrs 54 minutes.
Thus, a satellite takes 23 hours and 54 minutes to complete one revolution around the earth.
APPEARS IN
संबंधित प्रश्न
Why are geostationary satellites not useful for studies of polar regions?
Complete the following table.
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?
Considering first correlation, complete the second.
Hubble telescope : At 569 km above the earth’s surface
Orbit of Hubble telescope : .............................
Mahendra and Virat are sitting at a distance of 1 metre from each other. Their masses are 75 kg and 80 kg respectively. What is the gravitational force between them? G = 6.67 x 10-11 Nm2/kg2
The function of a satellite launcher is based on Newton's second law of motion.
Distinguish between:
High Earth orbit - Medium Earth orbit.
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 |
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.
The orbital velocity of the satellite depends on its ______.
Give scientific reasons.
The geostationary satellite is not useful in the study of polar regions.
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 ______.
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.
Complete the following equations:
