# I dont get this

1. Jul 23, 2009

### mvk1

ALl the books say that g is constant near the earth's surface but how is this even possible if it is dependant on the distance from the centre of the earth? because it depends on where we are why does it stil remain constant?

2. Jul 23, 2009

### mikelepore

Since the surface of the earth is about 6400 kilometers from the center of the earth, that distance changes by a very small percentage if you go up and down small hills, therefore g changes only by a very small percentage. But g isn't absolutely constant. Boulder, Colorado, highest altitude in the U.S., has a smaller g than Death Valley, California, lowest altitude in the U.S. But g in both place is the same to a few decimal places of precision.

I saw this is a book somwhere:
g at sea level: 9.808 m/s^2.
g at an altitude of 16 kilometers: 9.757 m/s^2.

Last edited: Jul 23, 2009
3. Jul 23, 2009

### sylas

g also varies by a tiny amount depending on local concentrations of mass. The "astronomy picture of the day" for Nov 13, 2001 is a graphical view of variation in g over the Earth.

4. Jul 23, 2009

### maverick_starstrider

5. Jul 23, 2009

### JazzFusion

1. The force due to gravity between two objects is Fg = (m1 x m2 x G) / r2,
where m1 is the mass of one object, m2 is the mass of the second, G is the gravitational constant, and r is the distance between them.

2. The force due to the gravity of the Earth on an object of mas m is Fge = (m x Me x G) / r2, where M is the mass of the Earth and r is the distance to the center of the Earth.

3. Since F = m a , the acceleration due to gravity from the Earth on that mass is (M x G) / r2.

4. The mean radius of the Earth is about 6371 km. The height of Mt Everest is 8.85 km. Therefore, 'r' at sea level is about 6371 km, while 'r' on the top of Mt Everest is about 6380 km.

5. Substituting these values for 'r', (along with M = 5.974 x 1024 kg and G = 6.674 x 10-11 m3/kgs2), you can calculate the acceleration due to gravity at sea level and on the top of Mt Everest (presumably, anywhere on Earth would have 'r' somewhere between those two values). The difference between the two is less than .3% (which any Physics student I nkow in a PHY100 lab would be happy to get!).

So, for all practical purposes, you can assume the acceleration due to gravity is constant on or near the surface of the Earth. If you want to launch a satellite, or lob a missile with enough precision to hit a specific window in an al Qaeda hideout, then you will need more precision.

ADDED ON EDIT: Man, you guys type too fast!! THREE answers while I was typing!!

6. Jul 26, 2009

### SW VandeCarr

Centrifugal "force" also reduces the effect of gravity at the equator. As you move toward either pole, this effect diminishes. It's less than 1%, but it's generally stronger than local gravitational anomalies.

http://en.wikipedia.org/wiki/Earth's_gravity

Last edited: Jul 26, 2009
7. Jul 26, 2009

### Bob S

Last edited by a moderator: Apr 24, 2017
8. Jul 27, 2009

### Nabeshin

Aside from what everyone is saying about how near the earth's surface implies a relatively low $$\frac{\Delta r}{r}$$, there is another (I think cooler) way to think about this.

The thing to note is that near the earth's surface, the earth appears to be flat and infinite. If you consider gravity in a gaussian sense (the same way the electrical field is treated in introductory physics), a flat, infinite sheet of mass produces a uniform gravitational field!

9. Jul 27, 2009

### AUMathTutor

Wow, that's some good outside-the-box thinking, Nabeshin. You're right, that way is cooler. It's like we're a little differential surface elements in somebody's flux calculation.

10. Jul 27, 2009

### SW VandeCarr

Yes, but despite the equatorial bulge and polar flattening, the measured sea level (net effective) acceleration of gravity at the equator is $$9.789 ms^{-2}$$ and $$9.832 ms^{-2}$$ at the north pole due to the earth's rotation.

http://www.space-electronics.com/Literature/Precise_Measurement_of_Mass.PDF see Abstract, latitude

Last edited by a moderator: Apr 24, 2017
11. Jul 27, 2009

### Hurkyl

Staff Emeritus
It's good thinking, but it's not outside-the-box.

12. Jul 27, 2009

Staff Emeritus
In the pre-GPS days, you could find something called "length of the seconds pendulum" tabulated in books. This was, as the name suggests, the length of a pendulum with a period of one second. The detail varied - latitude, elevation and longitude all have an effect. The amount of tabulated data was really remarkable.

(I don't suppose a Mentor could put a more descriptive title on this thread?)

13. Jul 27, 2009

### AUMathTutor

"It's good thinking, but it's not outside-the-box. "

Way to contribute to the thread, Hurkyl. That's some outside the box peanut-gallerying. Are we really going to argue about what constitutes an approach's being "outside the box"? If so, I'll retract the claim and save us all a lot of headaches.

Gravity at Earth's surface is fun. It always warms my heart to know that inquiring minds like mvk1's can use logic and a basic understanding of core physics to ask some really interesting questions. To summarize, he was right: g does vary, not by much, but it does. The "not by much" part is interesting to know and lends itself to a deeper understanding of the phenomenon.

Last edited: Jul 27, 2009
14. Jul 27, 2009

### Hurkyl

Staff Emeritus
The retraction isn't really believable if you argue immediately beforehand. :tongue:

This is not the sort of thing we want students thinking "wow, that's clever -- I wish I was brilliant enough to think of things in such a unique and clever way".

This is the sort of thing we want students thinking "That's a useful trick. I should start thinking about things in that way too. Oh wow, my textbooks were teaching me that all along!"