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
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
Name the first artificial satellite sent by Russia in space.
Distinguish between:
High Earth orbit - 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 |
| The highest height ever reached by man made balloon | 400 | 9.8 |
| Orbit of spacecraft | 35700 | 9.77 |
| Orbit of communication satellite | 0 | 8.7 |
Why are some satellites called geostationary?
What is microgravity?
Write down the formula of orbital velocity.
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:

