Weekly Challenge 7: Spaceship Earth

 Posted by DrJeff on September 4th, 2009

 Copyright 2009  |  About this blog



Photo caption: Computer-generated image of the Milky Way galaxy based on real data.


This post is a Dr. Jeff’s Weekly Challenge. It was requested by (teacher extraordinaire) Jami Lupold and her class in the great city of Houston, Texas, USA.

If you’re a teacher and your class has an idea for a blog post, slip me a note!


You wanted to be an astronaut? Poof. Done.


You and your friends are on a spaceship called Earth—with all known life aboard. With you sitting there calmly reading this, and no obvious need to hold on to something for dear life, it might seem that the spaceship under your feet is carrying you on a nice steady trajectory through space. Uh … Nope. Right now you’re being carried along on something more akin to a cosmic-sized amusement park ride. Earth is rotating on its axis, it’s orbiting the Sun, and the whole Solar System (the Sun and its planets, dwarf planets, asteroids, comets, and Trans-Neptunian Objects) is orbiting the center of the Milky Way galaxy. The Milky Way itself is moving relative to other nearby galaxies, the local group of galaxies is moving through a greater space, and all this is set against a backdrop of an expanding fabric of space and time across the entire universe. I know!!!! (© Craig Ferguson) Dizzy?


[Pleasant Ding] The captain has just turned on the seat belt sign. There may be some conceptual turbulence up ahead. But I’ll make the ride as smooth as possible.


OK, I think a Weekly Challenge requesting that you calculate all the spinning, and revolving, and free-flying is a bit much, so let’s concentrate on three things:


Here now the Challenge—


1. The Effect of Earth’s rotation: say you’re just standin’ there on Earth’s equator, all peaceful-like (a shout-out to my favorite class in Texas), minding your own business. How fast are you moving due to Earth’s rotation?

Hint: the Earth rotates once in 24 hours, and think “circumference of Earth.”


2. The Effect of Earth’s revolution (it’s orbit around the Sun):

Now you’re standing on the North Pole. Why? Well it’s a place where we can forget about the speed you’re carried due to Earth’s rotation. Up here (excuse me for a moment …. brrrrr) Earth’s rotation just gently rotates you once in 24 hours on the spin axis you’re standing on. But you’re still movin’ through space, yes you are … ’cause the entire Earth is zipping along in its orbit around the Sun. Your assignment #2 if you choose to accept it: how fast are you moving due to Earth’s motion around the Sun?

Hint: the Sun is 93,000,000 miles or 149,600,000 km (on average) from Earth, and it takes 1 year to go around once. This time think “circumference = 2 x pi x r”


3. The effect of Earth’s revolution around the center of the Milky Way galaxy:

This one is the toughest ’cause now I’ve relocated you to the north pole of … the SUN. Here, you don’t experience effects 1 and 2 above. From here you can just watch the spinning Earth as it orbits you. (Excuse me a minute …. hot, hot, hot!) What? Why did I put you on the NORTH POLE of the Sun? So I would not confuse you with your motion due to the Sun’s ROTATION.  (Hey, get with the program …. everything is in motion.)


Your assignment #3 is to figure out how fast the Sun (carrying the entire Solar System along for the ride) is moving in its orbit around the center of the Milky Way galaxy.

Hint: assume the Sun is 28,000 light years from the center, and completes 1 orbit in 240 million years

Oh, other needed Hint: 1 light year = 5.9 trillion miles or 9.5 trillion km

And yeah: think “circumference = 2 x pi x r” (again)


So get to figurin’, and remember that while you’re sittin’ there, you’re ZOOMING THROUGH SPACE due to those three effects you’re figurin’!


Thanks Jami and Jami’s class! Have a great school year!

—Doctor Jeff (Honorary Texan)


Answer now posted here!


Photo Credit: National Geographic Society Image Collection, 1999.



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One Response to “Weekly Challenge 7: Spaceship Earth”

  1. Benjamin D Brooks Says:
    September 13th, 2009 at 10:54 am

    Attempted Answers:
    1) speed of Earth rotation
    using D=VxT

    D =circumference
    =40,072,020m or 4.07202×10^7 m

    T= 24*60^2
    =86400s or 8.64×10^4 s

    therefore V = D/T
    =4.07202×10^7 / 8.64×10^4
    =4.638×10^3 m/s
    =4.64km/s <=ANSWER

    2) speed of Earth revolution
    using D=VT and C=2*pi*R
    where R=149,600,000km

    D=C=2 x pi x R

    therefore D=2x pi x 149,600,000,000 m
    =9.3996×10^11 m

    T= 31,556,926 (obtained from Google – could also do 365.25 x 86400)

    therefore V = D/T
    = 9.3996×10^11 / 31,556,926
    = 2.9786×10^4 m/s
    = 29.79km/s <=ANSWER

    3) speed of Solar revolution
    using D=VT and C=2*pi*R
    where R=28000 ltyr

    D=C=2 x pi x R

    D=2x pi x (28000 x (9.5×10^12 km) x 1000 m)
    =1.6713×10^22 m

    T=240 my X 31556926 (sec/yr)
    = 7.57336×10^15 s

    therefore V = D/T
    = 1.6713×10^22 / 7.57336×10^15
    = 2.206726×10^6 m/s
    = 2206.72×10^3 km/s <=ANSWER