Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
Answer:
R = 6.3456 10⁴ mile
Explanation:
For this exercise we will use Newton's second law where force is gravitational force
F = m a
The satellite is in a circular orbit therefore the acceleration is centripetal
a = v² / r
Where the distance is taken from the center of the Earth
G m M / r² = m v² / r
G M / r = v²
The speed module is constant, let's use the uniform motion relationships, with the length of the circle is
d = 2π r
v = d / t
The time for a full turn is called period (T)
Let's replace
G M / r = (2π r / T)²
r³ = G M T²²2 / 4π²
r = ∛ (G M T² / 4π²)
We have the magnitudes in several types of units
T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s
Re = 6.37 10⁶ m
Let's calculate
r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)
r = ∛ (1.027487 10²⁴)
r = 1.0847 10⁸ m
This is the distance from the center of the Earth, the distance you want the surface is
R = r - Re
R = 108.47 10⁶ - 6.37 10⁶
R = 102.1 10⁶ m
Let's reduce to miles
R = 102.1 10⁶ m (1 mile / 1609 m)
R = 6.3456 10⁴ mile
