Weightlessness*

Lawrence Davis

85 Weightlessness*

Uniform Circular Motion

We have seen that if the is found to be perpendicular to an object’s motion then it can’t do any on the object. Therefore, the net force will only change the object’s direction of motion, change it’s ) and the object must maintain a . The object will undergo , in which case we sometimes refer to the net force that points toward the center of the circular motion as the , but this is just a naming convention. The centripetal force is not a new kind of force, rather the centripetal force is provided by one of the forces we already know about, or a combination of them. For example, the centripetal force that keeps a satellite in orbit is just and for a ball swinging on a string in the string provides the .

A ball moves in a circle. The tangential velocity is indicated by an arrow perpendicular to the radius of the circle, in the direction the mass moves at the current moment. The centripetal force points toward the center of the circle. A second diagram shows the centripetal force is removed and the mass moves off in a straight line, in the direction it was moving the instant centripetal force was removed. — Left: A ball on a string undergoing circular motion with uniform (constant) speed. Right: The ball’s trajectory after the string breaks . Image Credit: Breaking String by Brews ohare via Wikimedia Common

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For both the ball and the satellite the points at 90° to the object’s motion so it can do no , thus it cannot change the of the object, which means it cannot change the of the object. How do we mesh this with , which says that objects with a net force must experience ? We just have to remember that acceleration is change per time and velocity includes and direction. Therfore, the constantly changing direction of constitutes a constantly changing velocity, and thus a constant, so all is good. Due to , we know that the points toward the center of the circular motion because that is where the points. As a result, that acceleration is called the . If the drops to zero (string breaks) the acceleration must become zero and the ball will continue off at the same speed in whatever direction it was going when the net force became zero (diagram on right above).

Centripetal Force and Acceleration

The size of the acceleration experienced by an object undergoing with radius $r$ at $v$ is:

(1) $\begin{equation*} a_c = \frac{v^2}{r} \end{equation*}$

Combined with we can find the size of the , which again is just the during :

(2) $\begin{equation*} F_{net} = ma = m \frac{v^2}{r} \end{equation*}$

Everyday Example: Rounding a Curve

What is the maximum that a car can have while rounding a curve with radius of 75 m without skidding? Assume the between tire rubber and the asphalt road is 0.7

First, we recognize that as the car rounds the curve at the must point toward the center of the curve and have the value:

$\begin{equation*} F_{net} = m \frac{v^2}{r} \end{equation*}$

Next we recognize that the only force available to act on the car in the horizontal direction (toward the center of the curve) is , so the in the horizontal direction must be just the frictional force:

$\begin{equation*} F_{f} = m \frac{v^2}{r} \end{equation*}$

We want to know the maximum speed to take the curve without slipping, so we need to use the maximum that can be applied before slipping:

$\begin{equation*} \mu_s F_N = m \frac{v^2}{r} \end{equation*}$

Notice that we have used static friction even though the car is moving because we are solving the case when the tires are still rolling and not yet sliding. Kinetic friction would be used if the tires were sliding.

For a typical car on flat ground the will be equal to the of the car:

$\begin{equation*} \mu_s mg = m \frac{v^2}{r} \end{equation*}$

Then we cancel the from both sides of the equation and solve for :

$\begin{equation*} v = \mu_s g r \end{equation*}$

Inserting our values for friction coefficient, g, and radius:

$\begin{equation*} v = sqrt{0.7(9.8\,\bold{m/s}) (75\,\bold{m})} = 22\,\bold{m/s} = 49\,\bold{MPH} \end{equation*}$

The inverted airfoils (wings) on a shape of a formula one race car create a downward aerodynamic force, increasing normal force from the road and thus friction as well. Image Credit: Italian V12 by Rob Oo via Wikimedia Commons

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Weightlessness

When you stand on a scale and you are not in , then the may not be equal to your and the weight measurement provided by the scale will be incorrect. For example, if you stand on a scale in an elevator as it begins to move upward, the scale will read a weight that is too large. As the elevator starts up, your motion changes from still to moving upward, so you must have an upward and you must not be in equilibrium. The from the scale must be larger than your weight, so the scale will read a value larger than your weight.

In similar fashion, if you stand on a scale in an elevator as it begins to move downward the scale will read a that is too small. As the elevator starts down, your motion changes from still to moving downward, so you must not be in , rather you have a downward . The from the scale must be less than your weight.

Taking the elevator example to the extreme, if you try to stand on a scale while you are in , the scale will be falling with the same as you. The scale will not be providing a to hold you up, so it will read your as zero. We might say you are weightless. However, your weight is certainly not zero because weight is just another name for the, which is definitely acting on you while you . Maybe normal-force-less would be a more accurate, but also less convenient term than weightless.

We often refer to astronauts in orbit as weightless, however we know the must be acting on them in order to cause the required for them to move in a circular orbit. Therefore, they are not actually weightless. The astronauts feel weightless because they are in along with everything else around them. A scale in the shuttle would not read their weight because it would not need to supply a to cancel their weight because both the scale and the astronaut are in free fall toward Earth. The only reason they don’t actually fall to the ground is that they are also moving so fast to their downward that by the time they would have hit the ground, they have moved sufficiently far to the side that they end up falling around the Earth instead of into it.

Everyday Example: Orbital Velocity

How fast does an object need to be moving in order to around the Earth (remain in orbit)? We can answer that question by setting the equal to the , given by Newton’s ( $F_g$ = mg is only valid for object near Earth’s surface, remember):

$\begin{equation*} F_g = \frac{Gm_1 m_2}{r^2} \end{equation*}$

Recognizing that is the in this case, and that $m_1$ is the Earth’s and $m_2$ is the orbiting object’s mass:

$\begin{equation*} \frac{Gm_1 m_2}{r^2} = m_2 \frac{v^2}{r} \end{equation*}$

Cancelling $m_2$ and one factor of $r$ from both sides and solving for :

$\begin{equation*} v= \sqrt{\frac{Gm_1}{r}} \end{equation*}$

We see that the necessary orbit speed depends on the radius of the orbit. Let’s say we want a low-Earth orbit at an altitude of 2000 km, or $2000\times 10^3\,\bold{m}$ . The radius of the orbit is that altitude plus the Earth’s radius of $6378 \,\bold{km}$ to get $r = 8378 \,\bold{km}$ or $r = 8378\times 10^3\,\bold{m}$ . Inserting that total radius and the gravitational constant, $G = 6.67408 \times 10^{-11}\,\bold{ m^3 kg^{-1}s^{-2}}$ , and the Earth’s mass: $m_1 = 5.972 \times 10^{24}\,\bold{kg}$ :

$\begin{equation*} v= \sqrt{\frac{(6.67408 \times 10^{-11}\,\bold{ m^3 kg^{-1}s^{-2}})(5.972 \times 10^{24}\,\bold{kg}) }{(8378\times 10^3\,\bold{m})}} \approx 7000 \,\bold{m/s} \approx 15,000 \,\bold{MPH} \end{equation*}$