Advertisements
Advertisements
प्रश्न
Numerical problem.
At an orbital height of 400 km, find the orbital period of the satellite.
उत्तर
h = 400 × 103m, R = 6371 × 103m,
v = 7616 × 103 kms– 1.
Substituting the values,
T = `"2π(R+h)"/"v"`
T = `6.28 × 6771/7616`
T = 5.583 × 103s = 5583 .
T ≈ 93 min
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?
Why are geostationary satellites not useful for studies of polar regions?
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?
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
Write functions of Military satellite and Navigational satellite.
Write down the formula of orbital velocity.
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.
Complete the Following table.
| Sr. no. | Type of Satellite | The names of Indian Satellite and launcher |
| (1) | Navigational Satellite | Satellite: ______ |
| Launcher: ______ | ||
| (2) | Earth Observation Satellite | Satellite: ______ |
| Launcher: ______ |
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.)
