Another Gravity Question. Surface Gravity This Time.

[BigDon]

So, if the Earth was squeezed down to the size of a meter, allowed to cool and then stood upon, would I experiance more surface gravity because more of the Earth was directly underneath me?

Or does it work that way?

(Hey, before coming here I never knew most meteor only hit at terminal velocity and that you couldn't have a stabile geosynchronous orbit on the Moon.)

[Jens]

The answer is, I'm almost positive, yes. I think we'd be talking pancake.

[Tobin Dax]

No, it's not because more of the Earth is directly beneath you. The larger surface gravity is due to the smaller radius.

A sphere (or spherical shell) with uniform density will act as if it were pulling on you from its center. The sideways pulls will be cancelled out by all of the other sideways pulls, and so everything together is pulling you straight down toward the center. This happens no matter the size of the sphere, whether it be 6,000 km or a meter (assuming that you are outside of it).

The larger surface gravity when standing on a one-meter Earth is due to how close you are to the center. Gravity follows an inverse-square law with distance; the force changes like 1/distance²). So if you double your distance from the center, the gravity you feel will be 1/4 as strong (1/2distance²). If you stand on an Earth with half of the radius, the surface gravity is 4 times stronger, at 1/3 the radius it's 9 times stronger, and so on.

With a one-meter (uniform sphere) Earth, the surface gravity is about 40 million million times stronger because you are over 6 million times closer to the center point (where all of the mass appears to the concentrated).

So, if the Earth was squeezed down to the size of a meter, allowed to cool and then stood upon, would I experiance more surface gravity because more of the Earth was directly underneath me?

Or does it work that way?

(Hey, before coming here I never knew most meteor only hit at terminal velocity and that you couldn't have a stabile geosynchronous orbit on the Moon.)

[BigDon]

Wow, thank you Tobin.

Didn't you mention once you were a teacher?

[Jeff Root]

Tobin is being rather persnickety in saying "No", the stronger gravity
isn't because more mass is directly under you. Reasonably persnickety.
I'll be unreasonably persnickety by saying that it isn't your distance from
the center, but your distance from the mass. The center very conveniently
represents the location of the mass in a single coordinate, as Newton
figured out and Tobin explains, so your distance from the center of the
mass is equivalent to your distance from the mass, for purposes of
calculating gravitational field strength and the like.

-- Jeff, in Minneapolis

[WayneFrancis]

So, if the Earth was squeezed down to the size of a meter, allowed to cool and then stood upon, would I experiance more surface gravity because more of the Earth was directly underneath me?

Or does it work that way?

(Hey, before coming here I never knew most meteor only hit at terminal velocity and that you couldn't have a stabile geosynchronous orbit on the Moon.)

If the Earth suddenly shrunk to 1m diameter and you where some how left 6,371km above the new surface the gravity would not change for you at all. You would start falling at about 9.8m/s towards this new surface accelerating faster and faster as you fell. If you could slow yourself down before you reached this new 1m ball that has a mass of 5.9736 × 10²⁴kg but you being 6,371,000 times closer means you'd feel about 40 trillion times heavier then you do now.

As I answer posts as I read through threads I didn't see Tobin Dax already did the calculations. Well at least I can be confident that my maths are correct as we got the same answers

[Tobin Dax]

No problem, Don. (Yes, I teach at a community college.)

Tobin is being rather persnickety in saying "No", the stronger gravity
isn't because more mass is directly under you. Reasonably persnickety.
I'll be unreasonably persnickety by saying that it isn't your distance from
the center, but your distance from the mass. The center very conveniently
represents the location of the mass in a single coordinate, as Newton
figured out and Tobin explains, so your distance from the center of the
mass is equivalent to your distance from the mass, for purposes of
calculating gravitational field strength and the like.

-- Jeff, in Minneapolis

Yes, I was, since I was quickly answering the question and sticking to surface gravity. Your answer and Wayne's answer are better answers, since they cover the more general case.

As I answer posts as I read through threads I didn't see Tobin Dax already did the calculations. Well at least I can be confident that my maths are correct as we got the same answers

Just a back-of-the-envelope calculation with a little fudge-factor for accuracy. Six squared is 36, so I called 6.4 squared 40.

[WayneFrancis]

No problem, Don. (Yes, I teach at a community college.)


Yes, I was, since I was quickly answering the question and sticking to surface gravity. Your answer and Wayne's answer are better answers, since they cover the more general case.



Just a back-of-the-envelope calculation with a little fudge-factor for accuracy. Six squared is 36, so I called 6.4 squared 40.

I'm all for "back-of-the-envelope" or as I call them "napkin" calculations. I'm just glad our rough calculations matched