Saber Posted January 14, 2023 Posted January 14, 2023 (edited) inside planets................not @ the center but lets say like here Shown in the picture Is the gravity less than on the surface as there is some mass above and also less mass below your feet Or is the gravity more because your are closer the the center of mass ? Thanx Edited January 14, 2023 by Saber
iNow Posted January 14, 2023 Posted January 14, 2023 Depends on from which side you take the measurement.
Genady Posted January 14, 2023 Posted January 14, 2023 If the planet is spherically symmetric then only the planet mass between the body and the planet center attracts the body. Outer part of the planet cancels. So, the short answer, less gravity. * "between the body and the planet center" means a sphere centered in the planet center 1
Genady Posted January 14, 2023 Posted January 14, 2023 1 hour ago, Genady said: the short answer, less gravity. Sorry, I need to clarify. This answer assumes that the density is the same throughout the planet depth. In reality it is not so, of course. The answer depends on how the density changes with depth and might be different at different depths. 1
exchemist Posted January 14, 2023 Posted January 14, 2023 3 hours ago, Saber said: inside planets................not @ the center but lets say like here Shown in the picture Is the gravity less than on the surface as there is some mass above and also less mass below your feet Or is the gravity more because your are closer the the center of mass ? Thanx Look up Newton’s shell theorem. That explains why the mass at a radius greater than the location of the red dot has no gravitational effect on it. 1
Lorentz Jr Posted January 14, 2023 Posted January 14, 2023 (edited) 2 hours ago, Genady said: Sorry, I need to clarify. This answer assumes that the density is the same throughout the planet depth. No, you had it right the first time. The derivation of the shell theorem that exchemist just mentioned applies to each (spherically symmetric) shell individually, regardless of its density. For constant density, the gravitational field is linear in the radius: [math]\displaystyle{a(r) = \frac{G m(r)}{r^2}}[/math] [math]\displaystyle{a(r) = \frac{G \rho (\frac{4}{3} \pi r^3)}{r^2}}[/math] [math]a(r) = G \rho (\frac{4}{3} \pi r)[/math] [math]a(r) = G \rho (\frac{4}{3} \pi R) \displaystyle{ \left(\frac{r}{R}\right)}[/math] [math]\displaystyle{a(r) = \frac{GM}{R^2}\left(\frac{r}{R}\right)}[/math] [math]\displaystyle{a(r) = g\left(\frac{r}{R}\right)}[/math] Edited January 14, 2023 by Lorentz Jr 1
Genady Posted January 14, 2023 Posted January 14, 2023 3 hours ago, Lorentz Jr said: The derivation of the shell theorem that exchemist just mentioned applies to each (spherically symmetric) shell individually, regardless of its density. This is right. So, why I thought the clarification is needed? Just consider the extreme case where the whole mass is near the planet center and the rest of the planet is massless. As the body goes from the surface toward the center it just gets closer to the same mass. The gravitational force increases. In the other extreme case where the whole mass is in the surface shell and the planet is empty inside, the body will become weightless as soon as it crosses the surface. Ergo, the answer depends on the density distribution with the depth. Also, take a case where the entire mass is in the shell halfway down. As the body gets closer to that depth the gravity increases, and when it crosses that shell the gravity drops to 0. Ergo, the answer depends on the depth. 1
studiot Posted January 14, 2023 Posted January 14, 2023 Genady is correct the density and distribution of the mass makes a difference. There is nothing wrong with your maths, but note you initial conditions. 4 hours ago, Lorentz Jr said: each (spherically symmetric) shell individually, regardless of its density. For constant density, the gravitational field is linear in the radius: However there is more to this than meets the eye since gravity, whether considered as force or as an acceleration is a vector, not a scalar. The distribition and density of the mass affects the direction of gravity. So the geometric centre and the centr of gravity no longer conincide. It is interesting to note the the discovery of the massiveness of the Himalaya was discovered by Everest from deductions on observations variations of alignment of plumb bobs on the plains beneath. Yet another effect has not been mentioned and is oft forgot. The reason why gravity is greater at the poles than the equator , due to centrigugal effects of our spinning planet. Once again this also has an effect on direction. 1
Lorentz Jr Posted January 14, 2023 Posted January 14, 2023 (edited) 1 hour ago, Genady said: the answer depends on the depth. You said "the short answer, less gravity" and "This answer assumes that the density is the same throughout the planet depth." So it looked like you were saying "less gravity assumes that the density is the same throughout the planet depth." The boundary between increasing gravity and decreasing gravity (i.e acceleration as a function of r) is constant gravity: [math]\displaystyle{a(r) = \frac{Gm(r)}{r^2} = g}[/math] [math]Gm = g r^2[/math] [math]G dm = 2 g r dr[/math] [math]\displaystyle{\rho(r) = \frac{dm}{dV} = \frac{dm}{ 4 \pi r^2 dr}}[/math] [math]\displaystyle{\rho(r) = \frac{dm/dr }{ 4\pi r^2 } = \frac{2gr}{4\pi r^2 G}}[/math] So [math]\displaystyle{\rho(r) = \frac{k}{r}}[/math], where [math]\displaystyle{k = \frac{g }{ 2 \pi G}}[/math]. In other words, the acceleration will be g all the way down as long as the density is inversely proportional to the distance from the center. It will decrease with r (ie. increase with depth) if the density depends on r more sensitively than 1/r, and will increase with r (decrease with depth) if it depends less. Edited January 14, 2023 by Lorentz Jr 1
Genady Posted January 14, 2023 Posted January 14, 2023 (edited) 28 minutes ago, Lorentz Jr said: it looked like you were saying... Fair enough. I should've said instead something like, "In that answer I've assumed..." Edited January 14, 2023 by Genady
Lorentz Jr Posted January 14, 2023 Posted January 14, 2023 13 minutes ago, Genady said: "In that answered I've assumed..." Okay, that makes sense. 1
Sensei Posted January 14, 2023 Posted January 14, 2023 https://www.google.com/search?q=gravity+anomaly+map https://en.wikipedia.org/wiki/Gravity_anomaly "The gravity anomaly at a location on the Earth's surface is the difference between the observed value of gravity and the value predicted by a theoretical model. If the Earth were an ideal oblate spheroid of uniform density, then the gravity measured at every point on its surface would be given precisely by a simple algebraic expression. However, the Earth has a rugged surface and non-uniform composition, which distorts its gravitational field." ps. Gravity anomaly is widely used in e.g. mining industry.
swansont Posted January 14, 2023 Posted January 14, 2023 4 hours ago, studiot said: Yet another effect has not been mentioned and is oft forgot. The reason why gravity is greater at the poles than the equator , due to centrigugal effects of our spinning planet. This has an effect on the normal force, which is what we perceive, but not on gravity.
Saber Posted January 16, 2023 Author Posted January 16, 2023 According to the shell theorem inside shells there is no gravity .....right or better say gravitational forces end up cancelling each other........... In a planet thats not a shell...................but @ the very very center...............can we assume its a thick thick shell ( thickness of the shell = R of the planet ) therefore @ the center of planets there is no gravity ........ ?
Lorentz Jr Posted January 16, 2023 Posted January 16, 2023 (edited) 4 hours ago, Saber said: In a planet thats not a shell...................but @ the very very center...............can we assume its a thick thick shell ( thickness of the shell = R of the planet ) therefore @ the center of planets there is no gravity ........ ? Yes, that's reasonable. I believe it would be an extension of Newton's original formulation for thin shells, but it's simple enough. You can think of any spherically symmetric planet as an onion, with lots of concentric shells. 🙂 Edited January 16, 2023 by Lorentz Jr 2
studiot Posted January 16, 2023 Posted January 16, 2023 On 1/14/2023 at 4:16 PM, swansont said: This has an effect on the normal force, which is what we perceive, but not on gravity. Yes indeed, strictly so. Also since the centrifugal force is imaginary, we really should instigate a more complicated analysis. However the effect still affects the direction of gravity since the rotation causing this apparent reduction in the magnitude of gravity is not in the plane of the gravitational attraction, so the effect depends upon latitude and the forces necessary to calculate the normal force must be added vectorially.
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