Advertisements
Advertisements
प्रश्न
Numerical problem.
At an orbital height of 400 km, find the orbital period of the satellite.
Advertisements
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
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 the proper name of the orbits of satellites shown in the following figure with their height from the earth’s surface.
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 |
Write a note on orbital velocity
How is a satellite maintained in nearly circular orbit?
Why are some satellites called geostationary?
What is microgravity?
Give scientific reasons.
The geostationary satellite is not useful in the study of polar regions.
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.)
