Shareef takes a look at his favorite scene from the trailer for Marvel's Black Panther, and he details the forces involved to create such a scene. Check out more videos and tutoring services at mathlooksgood.com
In the above video, astronaut Doug Wheelock explains that traveling in orbit requires a speed of 5 miles per second. In my previous post, we determined that an object needs to travel at approximately 8 km/s, or kilometers per second. How do we convert kilometers to miles to make sure that our calculations agree with those of an astronaut?
This is something that you can type into google to get the answer, but a teacher (or an exam) may require that you show your work. So let's get down!
Converting from one distance to another
When you convert from one unit to another, you may have to start with a basic conversion and work your way outward. For kilometers to miles, let's start with a basic conversion that everyone was taught in high school: 1 inch is equal to 2.54 centimeters. To properly use this, we need to convert out kilometers into meters and our inches into miles. We can then proceed as follows:
- There are 1,000 meters in a kilometer (the prefix "kilo-" means one thousand)
- There are 100 centimeters in a meter (the prefix "centi" means one hundred, as in century)
- There are 2.54 centimeter in an inch
- There are 12 inches in a foot
- There are 5,280 feet in a mile
We can use these facts to convert our value of 8 km/s by multiplying by each of the ratios given above. Note that we must take care that corresponding units are written diagonally from each other.
I color coded the units to emphasize that they must line up diagonally next to each other. This means that some conversion numbers will be in a numerator while others are in the denominator. For example, I need the km in the denominator of the second ratio to cancel out the km in the numerator in the first ratio.
Converting from one time to another
We can use the same idea to prove Wheelock's assertion that astronauts see approximately 16 sunrises and sunsets each day. This would mean that the ISS makes a complete revolution around the Earth approximately 16 times a day. Additionally, we need to use the following ratios:
- There are 60 seconds in an minute
- There are 60 minutes in an hour
- There are 24 hours in a day
Finally, we need to use a defined value for the distance around the Earth, also known as the circumference. This value is defined as 24,901 miles. So our final ratio is listed below.
- There is one revolution around the Earth every 24,091 miles
We will apply the same math as earlier, making sure that the numerator and denominators which are diagonal to each other have the same units so that they can cancel out.
According to our calculations, we can conclude that orbiting at 5 mi does cause you circle the planet approximately 16 times in one day.
On my science blog, I shared my general thoughts on the wonderful movie Hidden Figures. On this post, I want to answer the one question I've received the most - how do you even get something into orbit around the Earth? And how do you leave orbit entirely?
As with all of my posts, you can skip the diagrams if you want - the text should give you the general idea. Now on with the show!
Getting to Orbit
When we say a satellite is in orbit around the Earth, we mean that the satellite is traveling at the exact speed to counteract the force of gravity pulling it down. This means that the satellite is constantly falling, but its going fast enough so that it stays the same height above the Earth. This results in the satellite traveling in a circular motion. There are two forces acting on a satellite in this case:
- The centripetal force. For circular motion to occur, this force must point inward toward the center of the circle
- The gravitational force. This force exists between any two objects in our known galaxy. It is directed between the centers of both objects (in this case, the satellite and the Earth)
For our satellite to stay in a circle, both of these forces must be the same. This means that we can set the centripetal force equal to the gravitational force. Solving this equation gives us a velocity of approximately 8 kilometers per second, or 17,682 miles per hour. This is the speed that any satellite (including the International Space Station) has to achieve to be able to stay in orbit. It's also why satellite launches tend to curve right after leaving the launchpad - they need to get to the orbital speed and be parallel to the Earth's surface so they can be pulled into circular obit.
What if we wanted to find the speed needed to leave orbit entirely? This is what NASA did during our trips to the moon, and what NASA will do for our upcoming trips to Mars. The calculation is more complicated than getting to orbit, but we can find the speed needed by using a a familiar example of dropping a ball on the ground.
First, let's talk about energy. Total energy is expressed in two forms: kinetic energy (energy proportional to the speed of the ball) and potential energy (energy proportional to the distance above the surface). Before you drop a ball, it is motionless, so its kinetic energy is zero. Right before the ball impacts the ground, it is right next to the surface, so its potential energy is zero.
The total energy throughout this entire trip must remain the same, so we can set the the potential energy before dropping the ball equal to the kinetic energy right before it hits the ground. Solving this equation gives us a final velocity equal to the square root of 2 * times the acceleration due to gravity * the original height the ball was dropped from. Note that just like with getting to orbit, the speed does not depend on the mass of the ball.
The same energy situation applies if we throw a ball upwards, but our initial and final conditions are different. Now, we want to look at the energy of the ball just before it leaves the orbit of the Earth and compare it to the energy of the ball when it is an infinite distance off, when Earth's gravity diminishes to negligible value and the ball has zero energy. The reason we look at an infinite distance is because the force of gravity diminishes as you get farther from something, but never quite gets to zero.
Just like in the case of the dropped ball, we can set the potential and kinetic energy at the beginning (when the ball leaves orbit) equal to the energy when the ball stops. We can then solve for the escape velocity, which comes out to be very similar to the speed of the dropped ball! Solving this equation gives us a velocity of 11.2 kilometers per second, or 25,000 miles per hour.
Even though I grew up in northern New Jersey, I never went to the New Year’s ball drop in Times Square. Every year on top of the One Times Square building, a ball starts at the top of a pole at 11:59pm, and travels downward until it hits the bottom of the pole at 12pm. How much applied force is needed to make sure that it travels the length of the pole in the required time? Ignore friction and assume that the ball moves at a constant speed for the entire trip.
The ball, like everything else on Earth, has a force acting downwards due to gravity. We will choose to label this force as positive since it’s in the direction of motion. This force of gravity would cause the ball to fall too fast on its own. Thus, the force that we need is a force that acts upwards to slow the ball down just enough so that it travels the length of the pole in exactly one minute. We’ll choose to label this force as negative since it opposes motion.
We can find this force by using the concept of work. Work is defined as the energy transferred when applying a force over a distance. Both forces mentioned above transfer energy via work as the ball moves up and down the pole. The total work on the system is the
Total Work = Work due to gravity force – Work due to applied force
Total Work = Gravity force * distance of pole – Applied Force * distance of pole
The key to solving this problem is the work energy theorem, which states that the total work done on a system is equal the total change in the energy of a system as the ball travels from the top to the bottom of the pole. The energy can be broken down into kinetic energy (proportional to speed) and partially potential energy (proportional to height above ground).
Top of the pole – The ball has is moving so it has kinetic energy, and it also has potential energy
Bottom of the pole – The ball has the same kinetic energy as the top since it’s moving at a constant speed. The potential energy is lower because the ball is at a lower height.
Using this, we can also write our total work in terms of potential energy:
Total Work = Potential Energy at bottom of pole – Potential Energy at top of pole
We can then set our equations for Total work in 2 and 5 together and solve for our applied force.
If you’re interested in the actual math, the solution is below.
A pendulum is an object that hangs from a string - think of a swing hanging from a chain in a playground. If we pull the swing in any direction, the swing will ... well, swing to the other side. The swing will stop unless you are continually pushed or you swing your legs.
Describing the motion of that swing can get messy, so let's imagine an idealized swing where there is no friction and the weight of the chain is small enough to ignore. Also, we won't pull the swing too much, only a small angle from the original position (less than 15 degrees). Given those assumptions, we can get pushed once and swing forever.
This is known as a simple pendulum, and we can determine the speed and period (the amount of time it takes for one full swing) based solely on the length of swing chain.
In our simple pendulum, the swing seat does not move up or down the chain. This means that any forces that act up on the chain must equal any forces that act down on the chain.
- Fy, the force that acts down on the chain: The weight of the swing seat
- Since the full weight acts straight down, we're only concerned with a part of it
- Fc, the force that acts up on the chain: The force required for the swing seat to maintain a circular motion.
- This is always directed towards the center of the circle, so the force acts upward along the swing chain.
Calculations for the speed and period of the swing are given below.