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I think that, since Earth is spherical, the amount of mass in the upper half of Earth should be the same as the second half, therefore, if the strength of gravitational field increase linearly from distance 0 to R, it should also decrease at the same rate from R to 2R, however, it will never reach zero

Qmechanic
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Sherri
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    That graph isn't an accurate model of the Earth's gravity, it's for a simple uniform ball. See https://physics.stackexchange.com/a/18452/123208 for a more accurate graph. And also see https://en.wikipedia.org/wiki/Shell_theorem for the explanation of the simple graph in your question. – PM 2Ring May 29 '20 at 18:17
  • What do you mean by the “upper half”? – G. Smith May 29 '20 at 19:02
  • it should also decrease at the same rate from R to 2R, however, it will never reach zero It can’t do both of these; they are contradictory. – G. Smith May 29 '20 at 19:03
  • Can you explain the logic behind your “therefore”? I don’t see any connection between your premise and your conclusion. – G. Smith May 29 '20 at 20:34
  • @G. Smith If I draw a horizontal diameter along the center of the Earth, the part above this line is " the upper half" of the Earth, the portion below it will be " the bottom half" of the Earth – Sherri May 29 '20 at 21:54
  • @G. Smith by saying " it should also decrease at the same rater from R to 2R, however, it will never reach zero", I mean that the strength of gravitational field should decreases from R to 2R at the same rate as it increases from 0 to R, but, before it reach zero at 2R, the slope will somehow curved and keep going forever. – Sherri May 29 '20 at 21:55
  • @G. Smith And because mass is proportional to the strength of gravitational field, so if the mass in " the upper half" of the Earth and its bottom half are identical, thus the change of rate in gravitational field should also be the same, I mean like, it will decrease (from R to 2R ) at the same rate as it increase ( from 0 to R)... – Sherri May 29 '20 at 22:01
  • Sorry, I don’t see how considering the mass of the Earth to be divisible into a northern half and a southern half has anything to do with how it’s gravitational field should behave outside vs. inside. In any case, the diagram is correct. Inside the line is straight ($\propto r$). Outside it’s not ($\propto 1/r^2$). – G. Smith May 29 '20 at 22:09
  • @G. Smith If my logic doesn't work, then, what causes the change of rate in gravitational field so differently? Assuming there is a hole drilled through Earth from the North Pole to the South Pole, and a person starting to fall from the North Pole, he will first pass the crust, mantle, the outer core and then reach the inner core, so from this point he will keep falling but in the opposite path. Considering each layer as a whole and have different gravitational field, which the crust have the weakest gravitational field and the inner core have the strongest – Sherri May 29 '20 at 22:38
  • So then, I somehow come to a conclusion that the change of rate in gravitational field from 0 to R is the same as it is from R to 2R, the only difference is that one is increase and the another is decrease… – Sherri May 29 '20 at 22:38
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    From your comments I think that you are misunderstanding the horizontal axis of the graph. You seem to think that $0$ is at the surface (say, the North Pole), $R$ is at the center, and $2R$ is at the South Pole. Actually, $0$ is at the center, $R$ is at any point on the surface, and $2R$ is at any point 4000 miles up in the sky, twice as far from the center as the surface is. If this is not your misunderstanding, then I don’t understand what you are thinking. – G. Smith May 30 '20 at 00:19
  • thanks ，that is exactly what I was thinking – Sherri May 31 '20 at 01:05

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The mass in the upper half is not the same as the mass in the bottom half. If we assume uniform density then the mass for the center half would be $\rho.\frac{4}{3}\pi(\frac{1}{2}R)^3=\rho.\frac{1}{6}\pi R^3$ whereas the mass of the outer half would be $\rho.\frac{4}{3}\pi[R^3-(\frac{1}{2}R)^3]=\rho.\frac{7}{6}\pi R^3$.

The field is a straight line inside the earth because the mass is increasing as we move outwards. The mass, as a function of the distance from the centre is $M(r)=\rho \frac{4}{3}\pi r^3$ (keep in mind this is valid only for r < R where R is the radius of the earth). So the field is $F(r)=\frac{GM}{r^2}=\frac{G}{r^2}\frac{4\rho \pi r^3}{3}=\frac{4GM\rho \pi}{3}r$.

Once we've reached the distance R from the centre, there is no more mass being added so the force falls as the usual $1/r^2$.

baker_man
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  • Who told you the potential inside the ball was 4GMρπr/3? The potential inside a sphere is http://notizblock.yukterez.net/viewtopic.php?t=120#sphere.potential - also the graph doesn't show the potential, it shows the gravitational acceleration. By your logic the potential in the center would be 0, which would imply you need 0 energy to escape to infinity, which is clearly nonsense – Yukterez May 29 '20 at 18:34
  • @Yukterez oops, yeh I mixed up the potential and the field lol fixed it I think – baker_man May 29 '20 at 18:40