Advertisements
Advertisements
Question
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.
Advertisements
Solution
Given: Height of satellite (h) = 6400 km = 6.4 × 106 m
Velocity of satellite (Vc) = 5.6 km/s = 5.6 × 103 m/s
To find: Time taken for one revolution (T)
Formula: T = `(2π("R" + "h"))/"V"_"c"`
Calculation: T = `(2 xx 3.14 (6.4 xx 10^6 + 6.4 xx 10^6))/(5.6 xx 10^3)`
T = 14354 s ≈ 4 hours
APPEARS IN
RELATED QUESTIONS
Complete the following table.

Write the proper name of the orbits of satellites shown in the following figure with their height from the earth’s surface.

A group of students from COEP Pune sent a small satellite _______ through ISRO in 2016.
The function of a satellite launcher is based on Newton's second law of motion.
Distinguish between:
High Earth orbit - Medium Earth orbit.
Write a note on orbital velocity
How is a satellite maintained in nearly circular orbit?
Why are some satellites called geostationary?
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.
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 table.
| Sr. no. | Type of Satellite | The names of Indian Satellite and launcher |
| (1) | Navigational Satellite | Satellite: ______ |
| Launcher: ______ | ||
| (2) | Earth Observation Satellite | Satellite: ______ |
| Launcher: ______ |
Complete the following equations:

