How much faster would a person be moving if they stood on top of a 18.2 m high building compared to the ground? Take the period to be one day, and the radius of the earth to be 6400 km?
v = omega r
dv = omega dr
omega = 2 pi radians/(24*3600)seconds
dr = 18.2
dv = 2 pi (18.2) /(24*3600)
To determine how much faster a person would be moving when standing on top of an 18.2 m high building compared to the ground, we need to consider the rotational motion of the Earth.
The speed of an object in circular motion can be calculated using the formula:
v = r * ω
where:
v is the speed,
r is the radius, and
ω (omega) is the angular velocity.
Angular velocity, ω, can be calculated using the formula:
ω = 2π / T
where:
π is the mathematical constant pi (approximately 3.14159),
T is the period of rotation (in this case, one day).
Given that the radius of the Earth is 6400 km (or 6,400,000 m), we can calculate the angular velocity:
ω = 2π / (24 * 60 * 60) (since one day has 24 hours, 60 minutes, and 60 seconds)
Now, let's find the angular velocity (ω):
ω = 2π / 86400
= 7.272 x 10^(-5) rad/s (approximately)
Next, we can determine the speed of an object at the Earth's surface:
v = r * ω
= (6,400,000 + 18.2) * 7.272 x 10^(-5)
= 463.1128 m/s (approximately)
Therefore, a person standing on top of an 18.2 m high building would be moving approximately 463.1128 m/s faster compared to the ground.