Know the linear velocity, how to find the centripetal acceleration

Know the linear velocity, how to find the centripetal acceleration


It's simple:
We know V, ω
So the radius of circular motion r = V / ω
So centripetal acceleration:
a=R×ω^2=(v/ω)×ω^2=vω



The greater the angular velocity, the greater the linear velocity, the smaller the radius, and the greater the centripetal acceleration


A = (V ^ 2) / r = (ω ^ 2) r = ω V, according to several formulas, combined with what you know, you can see the magnitude of centripetal acceleration. When R is fixed, V is equivalent to ω, when a is large, ω is fixed, V is equivalent to R, when a is large, V is fixed, ω is large, etc



How to calculate that the linear velocity of a geostationary satellite is greater than that of an object at the equator of the earth


V = Δ L / Δ t l is the circumference. The L of the synchronous satellite is larger than that at the equator, but the time is equal, so it is correct



What are the characteristics of the linear velocity and angular velocity of the geostationary satellite launched at the earth's equator compared with the equator?


At the same angular velocity, the linear velocity of the satellite is 6 times that of the earth, because the distance between the geostationary satellite and the ground is 5 times the radius of the earth



The geostationary satellite is located above the equator, and its angular velocity with the earth,


The rotation speed of the synchronous satellite is the same as that of the earth
Because the radius of the satellite is larger than that of the earth, the linear velocity (v = ω T) is also larger than that of the earth



Geostationary satellite, near earth satellite, angular velocity of a point on the equator, period comparison
Don't use the specific value, just tell me the result and calculation formula,


Radius of the earth R
A point on the equator: angular velocity W = 2 π / 86400
Linear velocity v = wr
Period T = 2 π / W
Near earth satellite: v = sqrt (GM / R)
w=v/R
T=2π/w



What is the rotation speed of the earth on the equator


On the equator, the linear velocity of the earth's rotation is 40000 km / D, and the angular velocity is 360 ° / d



In the process of the earth's rotation, the angular velocity of each point on the ground is the same, but the linear velocity is from the pole (0km / D) to the equator (20000km / D)
What is the unit of km / D


Km = km d = day
Km / D is the unit of linear speed, i.e. km / d
However, the equatorial line speed should be 40000 km / D, not 20000 km / D
Supplement:
According to the Encyclopedia:
What is the unit of time
century
Year (y)
Month (m)
Week (W)
Day (d)
Hours (H)
Minutes (min)
Second (s)
Milliseconds (MS)
Microsecond (US)
Nanosecond (NS)
Picosecond (PS)
Femtosecond (FS)
The smallest is one femtosecond, which is the negative 15th power of 10, that is, one millionth of a second



The radius of the earth is about 6400km, and the angular velocity of objects on the earth's equator rotating with the earth,
During the flight performance, if an aircraft makes a uniform circular motion with a radius of 3000m and a linear speed of 150m / s, its cycle is


The angular velocity of the earth is 2 Π / 24 * 60 * 60 = 7.27 * 10 to the - 5th rad / s
The linear velocity of the earth is 2 Π R / 24 * 60 * 60 = 465.421m/s
Linear velocity = angular velocity * r
150 = angular velocity * 3000
Angular velocity = 0.05rad/s
Linear velocity = 2 * Π * r / T
150=2∏*3000/T
The period T is about 125.6637s
For reference only



It is known that the radius of the earth is 6400km, and a person with a mass of M = 60kg stands on the equator and makes a circular motion with the rotation of the earth. The linear velocity and the centripetal force required for him to make a circular motion are calculated. If the rotation speed of the earth is faster and faster, the object on the equator will float up when the angular velocity is ω. The first cosmic velocity is 7.9km/s, according to which the mass of the earth is estimated


Linear velocity v = 2 π R / T = 2 π * 6400000 / (24 * 60 * 60) = 465.42m/s
Centripetal force F = MV ^ 2 / r = 60 * 465.42 ^ 2 / 6400000 = 2n
MW ^ 2R = mg, w = 0.00124rad/s
GMM / R ^ 2 = MV ^ 2 / R: M = 7900 ^ 2 * 6400000 / (6.67 * 10 ^ - 11) = 5.98 * 10 ^ 24kg