A geostationary satellite is revolving the earth whose height from the earth's surface is 6 R.R is the radius of the earth.what will ne the time period if the height from the earth's surface is 3.5 R.
Answers
Answer:
12.37 hours
Prerequisites:
- Geostationary Satellite has a time period of 24 hours.
- It revolves along the speed of earth, therefore it does not change it's position from which it was placed.
Coming to your question,
According to Kepler's Third Law, Time Period of a planet around the sun is equal to cube root of the semi major axis.
Here, the planet is considered to be a satellite and the semi major axis is the distance between the earth and the satellite.
Interpreting the values,
⇒ T₁ = 24 hours ; T₂ = ?
⇒ R₁ = 6R + R = 7R ; R₂ = 3.5 R + R = 4.5 R
[ Here, R is calculated from center of earth therefore it is 7R and 4.5R respectively. ]
Formula = T² = ( R³ )
⇒ ( T₁ / T₂ )² = ( R₁ / R₂ )³
⇒ ( 24 / T₂ )² = ( 7 R / 4.5 R )³
⇒ 576 / T₂² = 343 / 91.125 [ R gets cancelled ]
⇒ 576 / T₂² = 3.76 [ approx. ]
⇒ T₂² = 576 / 3.76
⇒ T₂² = 153.19
⇒ T₂ = √ 153.19 ≈ 12.37 hours.
Hence the time period of the satellite at 3.5 R above the surface of Earth is 12.37 hours approximately