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College Physics by Openstax Chapter 6 Problem 29

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts

Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration 10g10g if the rider is 15.0 m from the center of rotation?

(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.34(b). At what angle θ\theta below the horizontal will the cage hang when the centripetal acceleration is  10g10g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle 10g10g should be.)

Figure 6.34 (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to always be along its axis.

Solution:

Part A

The centripetal acceleration, aca_c, is calculated using the formula ac=rω2a_c = r \omega ^2. Solving for the angular velocity, ω\omega, in terms of the other variables, we should come up with

ω=acr\omega = \sqrt{\frac{a_c}{r}}

We are given the following values:

  • centripetal acceleration, ac=10g=10(9.81 m/s2)=98.1 m/s2a_c = 10g = 10 \left( 9.81\ \text{m/s}^2 \right) = 98.1\ \text{m/s}^2
  • radius of curvature, r=15.0 mr = 15.0\ \text{m}

Substituting the given values into the equation,

ω=acrω=98.1 m/s215.0 mω=2.5573 rad/secω=2.56 rad/sec  (Answer)\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{98.1\ \text{m/s}^2}{15.0\ \text{m}}} \\ \\ \omega & = 2.5573\ \text{rad/sec} \\ \\ \omega & = 2.56\ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The free-body diagram of the force is shown

The free-body diagram of the rider’s cage that hangs on a pivot at the end of the arm of a large centrifuge. College Physics Problem 6-29
The free-body diagram of the rider’s cage hangs on a pivot at the end of the arm of a large centrifuge.

Summing forces in the vertical direction, we have

∑Fy=0Farmsin⁡θ−w=0Farm=wsin⁡θ Equation 1\begin{align*} \sum_{}^{} F_y & = 0 \\ \\ F_{arm} \sin \theta-w & = 0 \\ \\ F_{arm} & = \frac{w}{\sin \theta} \ \quad \quad \color{Blue} \text{Equation 1} \end{align*}

Now, summing forces in the horizontal direction, taking into account that FcF_c is the centripetal force which is the net force. That is,

Fc=mac\begin{align*} F_c & = m a_c \end{align*}

We know that FcF_c is equal to the horizontal component of the force FarmF_{arm}. That is Fc=Farmcos⁡θF_c = F_{arm} \cos \theta. Therefore,

Farmcos⁡θ=mac\begin{align*} F_{arm} \cos \theta & = m a_c \\ \end{align*}

Now, we can substitute equation 1 into the equation, and the value of the centripetal acceleration given at 10g10g. Also, we note that the weight ww is equal to mgmg. So, we have

Farmcos⁡θ=macwsin⁡θcos⁡θ=m(10g)mgcos⁡θsin⁡θ=10mg\begin{align*} F_{arm} \cos \theta & = m a_c \\ \\ \frac{w}{\sin \theta} \cos \theta & = m (10g) \\ \\ \frac{mg \cos \theta}{\sin \theta} & = 10 mg \\ \\ \end{align*}

From here, we are going to use the trigonometric identity tan⁡θ=sin⁡θcos⁡θ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We can also cancel mm, and g since they can be found on both sides of the equation.

1tan⁡θ=10tan⁡θ=110θ=tan⁡−1(110)θ=5.71∘  (Answer)\begin{align*} \frac{1}{\tan \theta} & = 10 \\ \\ \tan \theta & = \frac{1}{10} \\ \\ \theta & = \tan ^{-1} \left( \frac{1}{10} \right) \\ \\ \theta & = 5.71 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 6 Problem 27

The radius and centripetal acceleration of a bobsled turn on an ideally banked curve

Problem:

(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?

Solution:

Part A

For ideally banked curved, the ideal banking angle is given by the formula tan⁡θ=v2rg\displaystyle \tan \theta = \frac{v^2}{rg}. We can solve for rr in terms of all the other variables, and we should come up with

r=v2gtan⁡θr = \frac{v^2}{g \tan \theta}

We are given the following values:

  • ideal banking angle, θ=75.0 ∘\displaystyle \theta = 75.0\ ^\circ
  • linear speed, v=30.0 m/s\displaystyle v=30.0\ \text{m/s}
  • acceleration due to gravity, g=9.81 m/s2\displaystyle g=9.81\ \text{m/s}^2

If we substitute all the given values into our formula for rr, we have

r=v2gtan⁡θr=(30.0 m/s)2(9.81 m/s2)(tan⁡75∘)r=24.5825 mr=24.6 m  (Answer)\begin{align*} r & = \frac{v^2}{g \tan \theta} \\ \\ r & = \frac{\left( 30.0\ \text{m/s} \right)^2}{\left( 9.81\ \text{m/s}^2 \right)\left( \tan 75^\circ \right)} \\ \\ r & = 24.5825\ \text{m} \\ \\ r & = 24.6\ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The radius of the ideally banked curve is approximately 24.6 m24.6\ \text{m}.

Part B

The centripetal acceleration aca_c can be solved using the formula

ac=v2ra_c = \frac{v^2}{r}

Substituting the given values, we have

ac=v2rac=(30.0 m/s)224.5825 mac=36.6114 m/s2ac=36.6 m/s2  (Answer)\begin{align*} a_c & = \frac{v^2}{r} \\ \\ a_c & = \frac{\left( 30.0\ \text{m/s} \right)^2}{24.5825\ \text{m}} \\ \\ a_c & = 36.6114\ \text{m/s}^2 \\ \\ a_c & = 36.6 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal acceleration is about 36.6 m/s236.6\ \text{m/s}^2.

Part C

To know how large is the computed centripetal acceleration, we can compare it with that of acceleration due to gravity.

acg=36.6114 m/s29.81 m/s2=3.73\frac{a_c}{g} = \frac{36.6114\ \text{m/s}^2}{9.81\ \text{m/s}^2} = 3.73

The computed centripetal acceleration is 3.73 times the acceleration due to gravity. That is ac=3.73ga_c = 3.73g.

This does not seem too large, but it is clear that bobsledders feel a lot of force on
them going through sharply banked turns!

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College Physics by Openstax Chapter 6 Problem 26

The Ideal Speed on a Banked Curve

Problem:

What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

Solution:

The formula for the ideal speed on a banked curve can be derived from the formula of the ideal angle. That is, starting from tan⁡θ=v2rg\tan \theta = \frac{v^2}{rg}, we can solve for vv.

v=rgtan⁡θv = \sqrt{rg \tan \theta}

For this problem, we are given the following values:

  • radius of curvature, r=100 mr=100\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g=9.81\ \text{m/s}^2
  • banking angle, θ=20.0∘\theta = 20.0 ^\circ

If we substitute the given values into our formula, we have

v=rgtan⁡θv=(100 m)(9.81 m/s2)(tan⁡20.0∘)v=18.8959 m/sv=18.9 m/s  (Answer)\begin{align*} v = & \sqrt{rg \tan \theta} \\ \\ v = & \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right) \left( \tan 20.0 ^\circ \right) } \\ \\ v = & 18.8959\ \text{m/s} \\ \\ v = & 18.9\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ideal speed for the given banked curve is about 18.9 m/s18.9\ \text{m/s}.

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College Physics by Openstax Chapter 6 Problem 25

The ideal banking angle of a curve on a highway

Problem:

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

Solution:

The ideal banking angle (meaning there is no involved friction) of a car on a curve is given by the formula:

θ=tan⁡−1(v2rg)\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)

We are given the following values:

  • radius of curvature, r=1.20 km×1000 m1 km=1200 m\displaystyle r = 1.20\ \text{km} \times \frac{1000\ \text{m}}{1\ \text{km}} = 1200\ \text{m}
  • linear velocity, v=105 km/h×1000 m1 km×1 h3600 s=29.1667 m/s\displaystyle v=105\ \text{km/h}\times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} = 29.1667\ \text{m/s}
  • acceleration due to gravity, g=9.81 m/s2\displaystyle g = 9.81\ \text{m/s}^2

If we substitute these values into our formula, we come up with

θ=tan⁡−1(v2rg)θ=tan⁡−1[(29.1667 m/s)2(1200 m)(9.81 m/s2)]θ=4.1333∘θ=4.13∘  (Answer)\begin{align*} \theta & = \tan^{-1} \left( \frac{v^2}{rg} \right) \\ \\ \theta & = \tan^{-1} \left[ \frac{\left( 29.1667\ \text{m/s} \right)^2}{\left( 1200\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\ \theta & = 4.1333 ^\circ \\ \\ \theta & = 4.13 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ideal banking angle for the given highway is about 4.13∘ 4.13 ^\circ.

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College Physics by Openstax Chapter 6 Problem 24

Centripetal Force of a Rotating Wind Turbine Blade

Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

Solution:

We are given the following values:

  • radius, r=100 mr=100\ \text{m}
  • angular velocity, ω=0.5 rev/sec×2π rad1 rev=3.1416 rad/sec\omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}
  • mass, m=4 kgm=4\ \text{kg}

Centripetal force FcF_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity vv and has magnitude Fc=macF_c = m a_c which can also be expressed as

Fc=mv2ror Fc=mrω2F_c = m \frac{v^2}{r} \quad \text{or} \quad \ F_c = mr \omega^2

For this particular problem, we are going to use the formula Fc=mrω2F_c = mr \omega^2. If we substitute the given values, we have

Fc=mrω2Fc=(4 kg)(100 m)(3.1416 rad/sec)2Fc=3947.8602 NFc=4×103 N  (Answer)\begin{align*} F_c & =mr \omega^2 \\ \\ F_c & = \left( 4\ \text{kg} \right)\left( 100\ \text{m} \right)\left( 3.1416\ \text{rad/sec} \right)^2 \\ \\ F_c & = 3947.8602\ \text{N} \\ \\ F_c & = 4 \times 10^3\ \text{N}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal force on the end of the wind turbine blade is approximately 4×103 N4 \times 10^3\ \text{N}.

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College Physics by Openstax Chapter 6 Problem 23

The centripetal force of a child riding a merry-go-round

Problem:

(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.

Solution:

Part A

We are given the following values: m=22.0 kgm=22.0\ \text{kg}, ω=40.0 rev/min\omega = 40.0\ \text{rev/min}, and r=1.25 mr=1.25\ \text{m}. We are asked to solve for the centripetal force, FcF_c.

Centripetal force FcF_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity vv and has magnitude Fc=macF_c=m a_c, which can also be expressed as Fc=mv2rF_c = m \frac{v^2}{r} or Fc=mrω2F_c = m r \omega ^2. Basing from the given values, we are going to solve the problem using the formula

Fc=mrω2F_c = m r \omega ^2

First, we need to convert the angular velocity ω\omega to rad/sec\text{rad/sec} for unit homogeneity.

40 rev/min×2π rad1 rev×1 min60 sec=4.1888 rad/sec40\ \text{rev/min} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} = 4.1888\ \text{rad/sec}

Now, we can substitute the given values into our formula.

Fc=mrω2Fc=(22.0 kg)(1.25 m)(4.1888 rad/sec)2Fc=482.5162 NFc=483 N  (Answer)\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 1.25\ \text{m}\right) \left( 4.1888\ \text{rad/sec}\right)^2 \\ \\ F_c & = 482.5162\ \text{N} \\ \\ F_c & = 483\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let us convert the angular velocity to radians per second.

3.00 rev/min×2π rad1 rev×1 min60 sec=0.3142 rad/sec3.00\ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}}=0.3142 \ \text{rad/sec}

Now, we can substitute the given values

Fc=mrω2Fc=(22.0 kg)(8.00 m)(0.3142 rad/sec)2Fc=17.3750 NFc=17.4 N  (Answer)\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 8.00\ \text{m}\right) \left( 0.3142\ \text{rad/sec}\right)^2 \\ \\ F_c & = 17.3750\ \text{N} \\ \\ F_c & = 17.4\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For the first centripetal force we solved in Part A,

Fcw=483 N(22 kg)(9.81 m/s2)=2.24\frac{F_c}{w} = \frac{483\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 2.24

The centripetal force is 2.24 times the weight of the child.

For the centripetal force we solved in Part B, we have

Fcw=17.4 N(22 kg)(9.81 m/s2)=0.0806\frac{F_c}{w} = \frac{17.4\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 0.0806

The centripetal force is only about 8% of the child’s weight.

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College Physics by Openstax Chapter 6 Problem 21

Centripetal acceleration of an amusement park ride shaped like a Viking ship

Problem:

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. The speed at the bottom of the arc is 23.4 m/s.

(a) What is the centripetal acceleration at the bottom of the arc?

(b) Draw a free-body diagram of the forces acting on a rider at the bottom of the arc.

(c) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(d) Discuss whether the answer seems reasonable.

Solution:

Problem 6-20: The centripetal acceleration of the commercial jet’s tires, and the force of a determined bacterium in it

At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.

(a) At how many rev/min are the tires rotating?

(b) What is the centripetal acceleration at the edge of the tire?

(c) With what force must a determined 1.00×10−15 kg bacterium cling to the rim?

(d) Take the ratio of this force to the bacterium’s weight.

Solution:

We are given the following quantities: linear speed, v=60.0 m/s v=60.0 \ \text{m/s}, radius is half the diameter, r=0.425 m r=0.425 \ \text{m}.

Part A

We can compute the angular velocity based on the given using the formula, ω=vr \displaystyle \omega = \frac{v}{r}.

ω=vrω=60.0 m/s0.425 mω=141.1765 rad/sec\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{60.0 \ \text{m/s}}{0.425 \ \text{m}} \\ \\ \omega & = 141.1765 \ \text{rad/sec} \end{align*}

Now, we can convert this into the required unit of rev/min.

ω=141.1765 radsec×1 rev2π rad×60 sec1 minω=1348.1363 rev/minω=1.35×103 rev/min  (Answer)\begin{align*} \omega & = 141.1765\ \frac{\text{rad}}{\text{sec}} \times \frac{1\ \text{rev}}{2\pi\ \text{rad}} \times \frac{60\ \text{sec}}{1\ \text{min}} \\ \\ \omega & = 1348.1363 \ \text{rev/min} \\ \\ \omega & = 1.35 \times 10^{3} \ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The centripetal acceleration at the edge of the tire can be computed using the formula, ac=rω2 a_{c} = r \omega ^{2}.

ac=rω2ac=(0.425 m)(141.1765 rad/sec)2ac=8470.5918 m/s2ac=8.47×103 m/s2  (Answer)\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.425\ \text{m} \right) \left(141.1765\ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 8470.5918 \ \text{m/s}^2 \\ \\ a_{c} & = 8.47 \times 10 ^{3} \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

From the second law of motion, the force is equal to the product of the mass and the acceleration. In this case, we are going to use the formula, Fc=mac F_c = m a_c . We are given the mass to be m=1.00×10−15 kgm=1.00 \times 10 ^{-15}\ \text{kg} , and the centripetal acceleration is solved in Part B.

Fc=macFc=(1×10−15 kg)(8470.5918 m/s2)Fc=8.4705918×10−12 kg m/s2Fc=8.47×10−12 N  (Answer)\begin{align*} F_c & = ma_c \\ \\ F_c & = \left( 1 \times 10^{-15}\ \text{kg}\right) \left(8470.5918 \ \text{m/s}^2\right) \\ \\ F_c & = 8.4705918 \times 10 ^{-12}\ \text{kg m/s}^2 \\ \\ F_c & = 8.47 \times 10^{-12} \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The ratio of this force, Fc F_c to the weight of the bacterium is

Fcmg=8.4705819×10−12 N(1×10−15kg)(9.81 m/s2)Fcmg=863.4640Fcmg=863  (Answer)\begin{align*} \frac{F_c}{mg} & = \frac{8.4705819 \times 10 ^{-12}\ \text{N}}{\left( 1 \times 10^{-15} \text{kg} \right)\left(9.81 \ \text{m/s}^2 \right)} \\ \\ \frac{F_c}{mg} & = 863.4640 \\ \\ \frac{F_c}{mg} & = 863 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Problem 6-19: The angular velocity of an “artificial gravity”

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?

Solution:

We are given the following quantities:

radius=diameter2=200 m2=100 m\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
centripetal acceleration,ac=9.80 m/s2\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

ac=rω2a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

ω=acr\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

ω=acrω=9.80 m/s2100 mω=0.313 rad/sec  (Answer)\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\ \omega & = 0.313 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Problem 6-18: The linear speed of an ultracentrifuge and Earth in its orbit

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Solution:

Part A

We are given a linear speed of an ultracentrifuge of 0.50 km/s0.50\ \text{km/s}. We are asked to verify this value if we are given a radius of r=0.100 mr=0.100\ \text{m} and angular velocity of ω=50000 rev/min \omega = 50000 \ \text{rev/min}. We are going to use the formula

v=rωv = r \omega

Since we are given a linear speed in km/s\text{km/s}, we are going to convert the radius to km\text{km}, and the angular velocity to rad/sec\text{rad/sec}

r=0.100 m×1 km1000 m=0.0001 kmr=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km}
ω=50000 rev/min×2π rad1 rev×1 min60 sec=5235.9878 rad/sec\omega = 50000 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} =5235.9878\ \text{rad/sec}

Now, we can substitute these into the formula

v=rωv=(0.0001 km)(5235.9878 rad/sec)v=0.5236 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 0.0001 \ \text{km} \right)\left( 5235.9878 \ \text{rad/sec} \right) \\ \\ v & = 0.5236 \ \text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This value is about 0.500 km/s.

Part B

From Table 6.2 of the book

ParentSatelliteAverage orbital radius r(km)Period T(y)r3 / T2 (km3 / y2)
SunEarth1.496×1081.496 \times 10^{8} 13.35×10243.35 \times 10^{24}

Using the same formulas we used in Part A, we can solve for the linear velocity of the Earth around the sun. The radius is

r=1.496×108 kmr=1.496 \times10^{8} \ \text{km}

The angular velocity is

ω=1 revyear×2π rad1 rev×1 year365.25 days×1 day24 hours×1 hour3600 secω=1.9910×10−7 rad/sec\begin{align*} \omega & = 1 \ \frac{\text{rev}}{\text{year}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{year}}{365.25 \ \text{days}} \times \frac{1\ \text{day}}{24\ \text{hours}}\times \frac{1\ \text{hour}}{3600\ \text{sec}} \\ \\ \omega & = 1.9910 \times 10^{-7}\ \text{rad/sec} \end{align*}

The linear velocity is

v=rωv=(1.496×108 km)(1.9910×10−7) rad/secv=29.7854 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 1.496\times 10^{8}\ \text{km} \right)\left( 1.9910 \times 10 ^ {-7} \right) \ \text{rad/sec}\\ \\ v & = 29.7854\ \text{km/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The linear velocity is about 30 km/s.

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