An earth satellite moves in a circular orbit with a speed of 6.2km/s. Find the time of one revolution and its centripetal acceleration.
Answers
Step-by-step explanation:
Given An earth satellite moves in a circular orbit with a speed of 6.2
km/s. Find the time of one revolution and its centripetal acceleration.
- Speed = 6.2 km/s = 6200 m/s
- We know that vo = √GM / r
- Or r = GM / vo^2
- Also T = 2π r^3/2 / √GM
- = 2πGM / vo^3
- = 2 x π x 6.67 x 10^-11 x 6 x 10^24 / (6200)^3 x 60
- = 175 min
The time of one revolution and its centripetal acceleration are 1.05 × 10⁴ s and 3.7 m/s².
Step-by-step explanation:
(a) Time for one revolution
The time taken by the satellite to complete one revolution is given by the formula:
T = (2πr)/v → (equation 1)
On applying Newton's second law, we get,
F = ma
On applying Newton's gravitation law, we get,
F = (GmM)/r² = ma(rad)
⇒ a(rad) = v²/r
(GmM)/r² = m × v²/r
(GM)/r = v²
Where,
Gravitational constant = G = 6.67 × 10⁻¹¹ N.m²/kg²
Mass of earth = M = 5.97 × 10²⁴ kg
r = (GM)/v²
On substituting the values, we get,
r = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(6200)²
∴ r = 1.04 × 10⁷ m
On substituting the value of 'r' on equation (1), we get,
T = (2π × 1.04 × 10⁷)/(6200 × 60)
∴ T = 176 minutes
(b) Centripetal acceleration:
a(rad) = v²/r
On substituting the values, we get,
a(rad) = (6200)²/(1.04 × 10⁷)
∴ a(rad) = 3.7 m/s²