Advertisements
Advertisements
Question
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?
Advertisements
Solution
Given:
Height of the satellite, h = 35780 km
Let the original mass of Earth be M. Then its new mass will be 4M.
The time taken by the satellite to revolve around the Earth's orbit is given as
`T = (2piR)/(v_c)`
Now, vc is given as
`v_c = sqrt((GM)/(R + h))`
Thus,
`T = (2pi(R + h))/(sqrt((GM)/(R + h))) = (2pi(R + h)sqrt(R + h))/(sqrt(GM))` ...(i)
= `T propto 1/(sqrt(M))` ...(ii)
Thus, from equation (ii), we see that when the mass of the Earth becomes 4 times, the time period of revolution of the satellite should be halved.
i.e., `T_(4M) = T/2` ...(iii)
Now, h = 35780 km
M = 6 × 1024 kg
R = 6.4 × 105 m
G = 6.67 × 10−11 N m2/kg2
Putting the values of h, M and R first, we get
T = 24 h
Using (iii), we get
`T_(4M) = 24/2 = 12`h
RELATED QUESTIONS
Fill in the blank and explain the statement with reasoning:
If the height of the orbit of a satellite from the earth surface is increased, the tangential velocity of the satellite will ......
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?
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.
The function of a satellite launcher is based on Newton's second law of motion.
What is Medium Earth Orbit?
Distinguish between:
High Earth orbit - Medium Earth orbit.
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.
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 |
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]
What is microgravity?
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.
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.
Complete the following equations:
