An electron in an atom has a speed of 2.2 × 10^6 m/s and orbits the nucleus at a distance of 5 × 10^-11m.a)What is its centripetal acceleration? b)A neutron star of radius 17 km is found to rotate at 5 revolutions per second. What is the centripetal acceleration of a point on its equator?
To find the centripetal acceleration in both scenarios, we can use the formula:
Centripetal acceleration (a) = (v²) / r
Where:
a = centripetal acceleration
v = speed of the object
r = radius of the circular path
a) For the electron in an atom:
Given:
v = 2.2 × 10^6 m/s
r = 5 × 10^-11m
Substituting the values into the formula:
a = (2.2 × 10^6 m/s)² / (5 × 10^-11m)
In order to solve this calculation, we can follow these steps:
1. Square the velocity: (2.2 × 10^6 m/s)² = 4.84 × 10^12 m²/s²
2. Divide the squared velocity by the radius: (4.84 × 10^12 m²/s²) / (5 × 10^-11m)
3. Using the rules of scientific notation, divide the coefficients and subtract the exponents: 9.68 × 10^23 m/s²
Therefore, the centripetal acceleration of the electron is 9.68 × 10^23 m/s².
b) For the rotation of a point on a neutron star:
Given:
v = 5 revolutions per second (rev/s)
r = 17 km (which we will convert to meters)
First, let's convert the radius from kilometers to meters:
17 km = 17 × 10^3 m
Substituting the values into the formula:
a = (v²) / r
Convert the revolutions per second (rev/s) to meters per second (m/s):
1 revolution = 2π radians
5 rev/s = 5 × 2π rad/s
Now substitute the values:
a = ((5 × 2π rad/s)²) / (17 × 10^3 m)
In order to solve this calculation, we can follow these steps:
1. Square the angular velocity: (5 × 2π rad/s)² = 100 π² rad²/s²
2. Divide the squared angular velocity by the radius: (100 π² rad²/s²) / (17 × 10^3 m)
3. Using the value for π (approximately 3.14), divide the coefficients and subtract the exponents: 5.89 × 10^(-2) m/s²
Therefore, the centripetal acceleration of a point on the equator of the neutron star is 5.89 × 10^(-2) m/s².