Advertisements
Advertisements
प्रश्न
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]
Advertisements
उत्तर
T = `"2π(R+h)"/"v"`
86400 =` "2 × 3.14 × (6370 + 36000)"/"v"`
v = 6.28 × 42370
266083.6 km/sec
S = `"d"/"t"`
= `266083/24`
= 11086.79 km/h
APPEARS IN
संबंधित प्रश्न
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
Write the proper name of the orbits of satellites shown in the following figure with their height from the earth’s surface.
Name the first artificial satellite sent by Russia in space.
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 a note on orbital velocity
Numerical problem.
At an orbital height of 400 km, find the orbital period of the satellite.
Give scientific reasons.
The geostationary satellite is not useful in the study of polar regions.
Geostationary satellites are not useful for studies of the polar region.
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.)
