Suppose the surface (radius = r) of the space station in the figure is rotating at 38.6 m/s. What must be the value of r for the astronauts to weigh one-half of their earth-weight?
Require that
V^2/r = g/2
Solve for r.
To find the value of r for the astronauts to weigh one-half of their earth-weight, we can use the concept of centrifugal force.
Centrifugal force is the outward force experienced by an object when it is in a rotating reference frame. It is given by the formula: F = mω²r, where F is the centrifugal force, m is the mass of the object, ω (omega) is the angular velocity, and r is the radius.
In this case, the centrifugal force must be equal to half of the force of gravity on Earth, which is the weight of the astronauts. So we can set up the equation:
F = (1/2)mg
Where F is the centrifugal force, m is the mass of the astronauts, and g is the acceleration due to gravity on Earth.
We can express the angular velocity ω in terms of the surface velocity v and the radius r. The formula is given by: ω = v/r.
Substituting ω into the equation, we have:
(mv²/r) = (1/2)mg
Canceling out mass (m) from both sides of the equation, we get:
v²/r = (1/2)g
Now, rearrange the equation to solve for r:
r = 2v²/g
Substituting the given value v = 38.6 m/s and the acceleration due to gravity on Earth g = 9.8 m/s², we can calculate the value of r:
r = 2(38.6²) / 9.8
r = 1503.94 meters
Therefore, the value of r for the astronauts to weigh one-half of their earth-weight is approximately 1503.94 meters.