Three Centripetal Force Equations

## **Problem**

A **centripetal force** is a force that makes an object follow a curved trajectory. At every point on the curved trajectory, the centripetal force points in the direction perpendicular to the motion of the object. For the case of **uniform circular motion**, where the object moves in a circular trajectory at constant speed (tangential to the circular trajectory), the centripetal force is the net force that tends towards the center of the circle. The magnitude of the centripetal force

F
c

{\displaystyle F\_c}

 is

F
c

=

m

v
2

r

{\displaystyle F\_c = \frac{mv^2}{r}}

where

m

{\displaystyle m}

 is the mass of the object undergoing uniform circular motion,

v

{\displaystyle v}

 is the object's tangential speed, and

r

{\displaystyle r}

 is the radius of the circle (also called the **radius of curvature**).

**Part 1:** Express the tangential speed

v

{\displaystyle v}

 in terms of the radius of the circle

r

{\displaystyle r}

 and the period of one revolution

T

{\displaystyle T}

. Then use this expression of tangential speed to express the centripetal force in terms of the period.

**Part 2:** The frequency is defined as the reciprocal of the period

# f

1
T

{\displaystyle f = \frac{1}{T}}

. Express the centripetal force in terms of frequency.

**Part 3:** The Moon, weighing

7.348
×

10

22

{\displaystyle 7.348 \times {10}^{22} }

 kg, revolves around the Earth in a nearly circular orbit. The distance between the centers of the Earth and the Moon is roughly 384,000 kilometers. If it takes the Moon 27.32 days to orbit the Earth, what is the centripetal force of the Moon towards the Earth?

## **Solution**

**Part 1**

The tangential speed in uniform circular motion is the distance divided by the period. Since the distance is the circumference of the circle with radius

r

{\displaystyle r}

, then

# v

2
π
r

T

{\displaystyle v = \frac{2\pi r}{T} }

.
Thus,

F
c

=
m

(

2
π
r

T

)

2

(

1
r

)

{\displaystyle F\_c = m\left(\frac{2\pi r}{T}\right)^2 \left(\frac{1}{r}\right) }

F
c

=

m
4

π
2

r

T
2

{\displaystyle F\_c = \frac{m4\pi^2 r}{T^2} }

.

**Part 2**

Since

# f

1
T

{\displaystyle f = \frac{1}{T}}

 and

F
c

=

m
4

π
2

r

T
2

{\displaystyle F\_c = \frac{m4\pi^2 r}{T^2} }

, we get

F
c

=
m
4

π
2

r

f
2

{\displaystyle F\_c = m4\pi^2 r f^2 }

.

**Part 3**

The orbital period is 27.32 days, so select the formula in Part 1:

F
c

=

m
4

π
2

r

T
2

{\displaystyle F\_c = \frac{m4\pi^2 r}{T^2} }

.
Convert the length of the orbital period from days to seconds:

T

=
27.32

d
a
y
s

(

24

h
o
u
r
s

1

d
a
y

)

(

3600

s
e
c
o
n
d
s

1

h
o
u
r

)

=
2360448

s
.

{\displaystyle
\begin{align}
T &= 27.32 \; days \left(\frac{24 \; hours}{1 \; day} \right) \left(\frac{3600 \; seconds}{1 \; hour} \right) \\
&= 2360448 \; s.
\end{align}}

Convert the distance between the centers of the Earth and the Moon into meters:

r

=
3.84
×

10
5

k
m

(

1000

m

1

k
m

)

=
3.84
×

10
8

m

{\displaystyle
\begin{align}
r &= 3.84 \times 10^5 \;km \left(\frac{1000 \; m}{1 \; km} \right) \\
&= 3.84 \times 10^8 \; m
\end{align}}

Thus, the centripetal force is

F
c

=
(
7.348
×

10

22

)

4

π
2

(
3.84
×

10
8

)

2360448

2

=
2.00
×

10

20

N

{\displaystyle \begin{align}
F\_c &= (7.348 \times {10}^{22}) \frac{ 4\pi^2 (3.84 \times 10^8)}{{2360448}^{2}} \\
&= 2.00 \times {10}^{20} \; N
\end{align}
}