Skip to main content

Extra Credit Blog!!!

There are 7.6 billion people in the world. If everyoen gathered together at one place at one time and all jumped together, what woul happen to the Earth? 


Simply put, even if everyone on the planet jumped at the exact same time the eath would be virtually uneffected. This is becasue teh mass of the earth is approximately 5.972 * 10^624 kg and if you were to calculate the mass of all humans put together it would not come near that number. 

Conservation of momentum:
p1i + p2i = p1f + p2f
m1v1+m2v2=m1v1f+m2v2f

Numbers:
  • Human population as of December 2017= 7.6 billion
  • Mass of the Earth= 5.972 * 1-^24 kg
  • Average human mass=136 lb/62kg
  • Average vertical jump=0.5 m


In the equations, 1 represents the Earth and 2 represents the people.  In this situation the equation is elastic becuase the Eaerth is in the system and momentum is always conserved.

1/2m1v1^2+1/2m2v2^2=mgh  --> h initial = 0
1/2m1v1^2+1/2m2v2^2=0
1/2m1v1^2=-1/2m2v2^2
m1v1=-m2v2
v1=-m2v2/m1

Then I used this to calculate teh velocity of teh people, when h did not = 0. 

1/2m1v1^2+1/2m2v2^2=m2gh
m1v1^2+m2v2^2=2m2gh
m2v2^2=2m2gh-m1v1^2
v2^2=(2m2gh-m1v1^2)/m2
v2^2=2gh-m1v1^2/m2
v2=sq. rt 2gh-m1v1^2/m2

Then, I solved for the velocity of the earth using theses equations together:

v1^2=(2ghm2^2/m1^2)-(m2v1^2/m1)

Next, I solved for v1 to get the velocity of the Earth.

v1^2+(m2v1^2/m1)=(2ghm2^2/m1^2)
v1^2(1+m2/m1)=(2ghm2^2/m1^2)
v1^2=2ghm2^2/m1^2(1+m2/m1)
v1^2 = 2ghm2^2/ (m1^2+m2m1)
v1 = sq. rt. 2ghm2^2/ (m1^2+m2m1)

Finally, I then plugged in the values. First I found the mass of the total population of the people which was (7.6*10^9)*62=4.712*10^11kg.

v2 = sq. rt. 2*9.8* (4.712 x 10^11)^2 / (5.972 x 10^24)^2 + (5.972 x 10^24)*(4.712 x 10^11)
v2 = (2.17589 x 10^24)/ (3.56648 x 10^49)
v2 = 6.10095 x 10^-26 m/s

This means that the velocity of the jump would only be effected by 6.10095 * 10^-26 m/s. The total mass of the population is insignificant compared to the mass of the Earth. This makes it difficult for the jump to effect the Earth's momentum. Realistically people can only jump up to half a meter high. The increase of mass to the center of the Earth would increase the rate of rotation. Huge earthquakes only increase the earths rotation by 100,000ths of a second. However, the current population of the earth would have to increase by seven million times what it is now to eave make that much of a difference. The collective mass of the population is just too insignificant compared to the Earths so it will not have an effect. The T.F Green Airport in Warwick could handle thousands of passangers a day at a rate of 500% capacity for years without even making a dent in the crowd. 

Comments

Post a Comment

Popular posts from this blog

Physics of Black Holes...Or Lack Thereof

Isabella Jacavone To comprehend how the universe works, we must dwell into the most basic building blocks of existence; matter, energy, space, and time. NASA's  Physics of the Cosmos program involves cosmology, astrophysics, and fundamental physics intended to answer questions about the elusiveness of complex concepts such as black holes, neutron stars, dark energy, and gravitational waves. In this blog post, I'd like to elaborate on a subject that is very intriguing  to me; Black holes. And more specifically, what would happen if we got near one. A black hole is anything but a hole, but rather an immense amount of matter compacted into an extremely small area. A black hole is caused when, hypothetically, a star four times more massive than our sun collapses into a sphere no bigger than 600 square km. To put that in perspective, that's about the size of New York City. B lack holes were predicted by Einstein's theory of general relativity, which showed that when a
Post a Comment
Read more

Physics Behind Drone Flight

A drone flies by using its downward thrust and forcing air in a particular direction in order to sustain a certain speed as well as a specific height. In this video my friend and I had been flying a drone at exactly 4 mph which converts to 1.788 m/s. In this project, we will be determining the forces acting upon the drone in order to sustain a consistent flight in terms of velocity and height while excluding the effects of air resitance. The drone is flying at an angle of 28˚, this is found by extending the tilted axis of the drone to the horizontal and finding the angle with a protractor. From this angle we will be able to calculate the downward thrust and the acceleration of the drone that allows it to maintain its height and velocity during flight. When the mass of the drone is taken it results in 734 grams or .734 kilograms, which will also be used for the calculations within the project. The freebody diagram pictured above will alow us to derive the force equations f
Post a Comment
Read more

Aerodynamics of a Golf Ball

One may wonder how a small golf ball can travel at incredibly high speeds for such long distances.  While the swing of the club is a major component, the structure of the golf ball is quite important.  Unlike a baseball or tennis ball, a golf ball has dimples all over it (usually 336 dimples).  These dimples allow the golf ball to travel without facing much air resistance.  This diagram shows how air travels around the golf ball. The dimples on the golf ball also prevent drag that would occur in the wake region, resulting in further distance.  Also due to the contact with the club during the swing, the golf ball has backspin during its entire flight.  This diagram shows the motion of the golf ball mid flight with the lift force of F. There are hundreds of different types of golf balls that a player can choose.  Some show little affect to a player's game while others can alter their performance completely.  Personally, I prefer Callaway Supersoft golf balls, but it is entirely
Post a Comment
Read more