A neutron star has 12 km radius and rotation period 1.6 s. What's the centripetal acceleration on its surface at the equator?
a=w^2 r=(2PI/periodinSeconds)^2 * radiusinmeters
So your saying that:
2pi/1.6s^2*12000m would give the correct answer?
To find the centripetal acceleration on the surface of a neutron star at the equator, we can use the formula for centripetal acceleration:
centripetal acceleration = (angular velocity)*(radius)
The angular velocity is the rotational speed of an object, measured in radians per second (rad/s). The rotation period is given in seconds, so we need to convert it to radians per second. Since one full revolution is equal to 2π radians, we can convert the rotation period as follows:
angular velocity = (2π radians) / (rotation period)
Substituting the given values:
rotation period = 1.6 s
angular velocity = (2π) / (1.6) rad/s = 1.25π rad/s
Next, we can substitute the angular velocity and the radius into the formula for centripetal acceleration:
centripetal acceleration = (1.25π rad/s) * (12 km)
Here, it is important to convert the radius from kilometers to meters, as the unit of acceleration is meters per second squared (m/s²):
12 km = 12,000 m
centripetal acceleration = (1.25π rad/s) * (12,000 m)
To get the result, we can calculate this expression using a calculator, considering that π is approximately 3.14159:
centripetal acceleration ≈ 47,123.85 m/s²
Therefore, the centripetal acceleration on the surface of the neutron star at the equator is approximately 47,123.85 meters per second squared (m/s²).