PDA

View Full Version : Acceleration of Surface Gravity equation



Vermonter
2005-Jan-27, 08:38 PM
I know the equation for the acceleration of gravity near Earth's surface, but I have a question. My professor posed this to me: What would the surface acceleration of gravity be on a planet twice as massive as Earth, but half the diameter? I did some calculations...

G(2M)
-------
(1/2R)^2 this gave me 78.7 m/sec^2 this is about 8 times normal surface gravity. Is this correct?

A Thousand Pardons
2005-Jan-27, 08:54 PM
Looks to be exactly 8 times. 'Course, earth gravity varies a little.

If it had twice the mass, and the same average density, then its radius would be the cube root of two times R, so the gravity would be the cube root of two times the gravity.

swansont
2005-Jan-27, 08:55 PM
One might predict it to be exactly 8 times larger. :) A factor of two for the mass, and a factor of four for the radius (since it's squared)

jfribrg
2005-Jan-27, 09:19 PM
I agree about the 8g.
And of course 9.81 * 8 = 78.5 (rounded to significant digits), which is remarkably close to Vermonter's answer. I'm not sure where the 78.7 came from, unless an intermediate result was rounded [-X .

Normandy6644
2005-Jan-27, 09:57 PM
Of course, for people without calculators, g=10. :lol:

jfribrg
2005-Jan-27, 11:20 PM
Of course, for people without calculators, g=10. :lol:

In which case, the calculated answer would be 80. :-k

A Thousand Pardons
2005-Jan-28, 10:44 AM
I agree about the 8g.
And of course 9.81 * 8 = 78.5 (rounded to significant digits), which is remarkably close to Vermonter's answer. I'm not sure where the 78.7 came from, unless an intermediate result was rounded [-X .
Watch that finger! someone could lose an eye :)

If you use google's values for the gravitational constant, mass and radius of the earth, and calculate GM/r^2, you get less than 78.4

But they use the equatorial radius of the earth, and the polar radius is over 20 kilometers shorter. Besides, the acceleration varies by a half percent over the surface of the earth anyway. And just using 6x10^24 kg (rounded up from google's figure of 5.9742 x 10^24) gives you the 78.7 figure.

Eta C
2005-Jan-28, 01:54 PM
The discussions of local anomalies and the earth's oblateness are going into unnecessary detail here. I think swansont had the idea the question was aiming at: double the mass, half the radius = 8g at the surface. The actual numerical value is irrelevant to understanding the underlying concept that Newton's law of universal gravitation is inverse square in radius and linear in mass. If I were grading this course I would not require any numerical calculation of the value in m/sec^2 and accept 8g as the answer. This answer is the most general and works for any planet, moon, etc. (assuming the local value of g for each body, of course), not just the earth.

A Thousand Pardons
2005-Jan-28, 04:59 PM
If I were grading this course I would not require any numerical calculation of the value in m/sec^2 and accept 8g as the answer.
Sure (http://www.badastronomy.com/phpBB/viewtopic.php?p=406426#406426), but would you accept 78.7 m/s/s ? :)