Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
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?
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
What is Medium Earth Orbit?
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 |
The orbital velocity of the satellite depends on its ______.
Geostationary satellites are not useful for studies of the polar region.
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:
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.)
