Jump to content


Photo
- - - - -

Tangential speed at equator.


  • Please log in to reply
6 replies to this topic

#1 Gareth56

Gareth56

    Atom

  • Senior Members
  • 271 posts

Posted 20 January 2009 - 03:25 PM

In College Physics 7th Ed by Serway it states that the tangential speed of the surface of the Earth at the equator is 1700m/s!!

Is this correct? I thought it was more like 465m/s.
  • 0

#2 D H

D H

    Physics Expert

  • Resident Experts
  • 3,631 posts
  • LocationHouston, Texas

Posted 20 January 2009 - 03:37 PM

Can you give a quote from that book in context? You are correct: 2*pi/sidereal day * 6378 km = 465 m/s.
  • 0

#3 timo

timo

    Scientist

  • Senior Members
  • 3,205 posts
  • LocationGermany

Posted 20 January 2009 - 03:38 PM

 v = \frac{2 \pi r}{24 \text{ h}} \approx \frac{6 \cdot 6 \cdot 10^3 \text{ km}}{24\text{ h}}
= \frac{36\cdot 10^3}{24} \frac{\text{km}}{\text{h}} = 1500 \frac{\text{km}}{\text{h}}.. So if "tangential speed" is defined the way I assumed, then it seems you are right. However, seeing that I did some approximations it is well possible that the actual value would be ~1700 km/h. In that case the college book might have screwed up the units (or you misread them).
  • 0

#4 D H

D H

    Physics Expert

  • Resident Experts
  • 3,631 posts
  • LocationHouston, Texas

Posted 20 January 2009 - 04:29 PM

However, seeing that I did some approximations it is well possible that the actual value would be ~1700 km/h. In that case the college book might have screwed up the units (or you misread them).

Good catch. Gareth56, double-check the units in the text. Is it kilometers per hour or meters per second? 2*pi/sidereal day * 6378 km = 1674 km/hr.

Note well: It is 2*pi/sidereal day, not 2*pi/24 hours. There is a slight difference, probably important only to the truly pedantic. A sidereal day is 23.9344696 hours long.
  • 0

#5 Gareth56

Gareth56

    Atom

  • Senior Members
  • 271 posts

Posted 21 January 2009 - 11:20 AM

Here is the exact quote from the book:-

"Why is the launch area for the European Space
Agency in South America and not in Europe?

Explanation:-

Satellites are boosted into orbit on top
of rockets, which provide the large tangential speed
necessary to achieve orbit. Due to its rotation, the surface
of Earth is already traveling toward the east at a
tangential speed of nearly 1700 m/s at the equator.
This tangential speed is steadily reduced further
north, because the distance to the axis of rotation is
decreasing. It finally goes to zero at the North Pole.
Launching eastward from the equator gives the satellite
a starting initial tangential speed of 1700 m/s,
whereas a European launch provides roughly half that
speed (depending on the exact latitude)."

  • 0

#6 swansont

swansont

    Shaken, not stirred

  • Moderators
  • 26,259 posts
  • LocationWashington DC region

Posted 21 January 2009 - 11:37 AM

I'm guessing it's the unit mistake, and should be km/hr.

You could contact the publisher and point that out. They might have an errata page online. Is there a URL given in the book?
  • 0

Minutus cantorum, minutus balorum, minutus carborata descendum pantorum                                   To shake my vodka martini, click the up arrow ^

I am not a minimum-wage government shill

My SFN blog: Swans on Tea                                                           

 

 

                                                                                                                     

 

 


#7 Gareth56

Gareth56

    Atom

  • Senior Members
  • 271 posts

Posted 21 January 2009 - 11:52 AM

I understand there's now an 8th edition so hopefully the proof reader spotted the typo.
  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users