# Tag Info

## Hot answers tagged newtonian-gravity

20

Great photo! Edit: My language is "sloppy" (I like talking physics in "lay person" terms so anybody can understand) but @dcmkee made really nice comment clarifying my answer for the more advanced people. Thanks @dcmkee! Since the plane is in a loop there is significant g's due to centripetal acceleration. The water was being accelerated upward$^{1}$ with ...

14

It doesn't actually have anything to do with the plane being upside down, or even changing from a vertical direction to a horizontal one. It's purely the vertical velocity that's at play here. Imagine water being thrown upward. You know what, imagine a fountain, a really big fountain. As soon as the water leaves the underground pump, it starts falling back ...

7

I posted an answer to a similar question here: http://physics.stackexchange.com/a/97978/1255 The wording there was pretty much asking for a broad description of the physics (which come down to bifurcation possibilities) up to, and including, the topology changes of the planet. The paper I linked to seems to be pretty much the state of the art on this ...

6

When an object comes within the Roche limit, it breaks up because of tidal stresses - the part closest to the earth feels a stronger gravitational attraction than the furthest part. Hence, the closest part will fall a little faster than the trailing parts. As a result, "disintegration" does not mean that the body will fly apart like a bomb. Instead, it ...

5

Think about the work-kinetic energy theorem, which states that the net work done on an object is equal to its change in kinetic energy: $$W_{net}=\Delta\mathrm{KE}.$$ You are right that when lifting an object of mass $m$ by a height $h,$ in a uniform gravitational field, the work you do is $W_{you}=mgh$ (assuming, as you said, that you're applying a force ...

5

It's actually not entirely true that the strength of the Earth's gravitational field decreases as a function of depth. It is true for certain regions in the Earth, but it's untrue for others because of the non-trivial dependence of the Earth's density on depth. To see what's going on, assume that the Earth is a sphere whose density is spherically ...

5

At the moment the picture was taken the plane, with mass $M$ was performing an inside loop, and was almost exactly inverted. It was moving at a speed $V$ in a vertical circle with radius $R$; both of these are chosen by the pilot as he execute the loop. The physics of circular motion requires that the plane experience a force towards the centre of the ...

4

No, these building are still tiny compared to earth's crust mass distribution. One would need to build whole mountain ranges to detect changes in earth gravity field with high precision instruments. And even those wouldn't changed earth orbit measurably because even a mountain range is tiny compared to the mass of the whole earth. However mountain ranges ...

3

Actual aircraft attitude (inverted with respect to the ground, in this case) is irrelevant. All that matters is that for the few moments long enough to pour the water and snap the picture, the aircraft is experiencing some positive g-load (pilot feels that he is pushed into his seat). The aircraft could be in a barrel roll or a loop. Either way, it is in ...

3

That equation applies for point sources, which the Earth technically is not. We can, however, treat the Earth as a point source as long as its internal structure is irrelevant (i.e. as long as we are outside of it). Once we enter the surface of the Earth, we can no longer simplify it by pretending it's a point and we have to perform a full analysis of the ...

2

This answer is an complement to Chris White's answer. Fist of there is no explicit equations for the position of an object following a Kepler orbit as a function of time. However, when the initial conditions are known, the path the object will follow can be found, as well as the velocity, acceleration, ect. at any given position. This path can be described ...

2

It seems you've done the hard part already, which is to evolve the object's position as a function of time. And moreover, the simulation seems stable over a number of orbits. (But eventually things start to go wrong; you may want to look at an answer I wrote to What is the correct way of integrating in astronomy simulations?) So my understanding is all you ...

2

The moon orbits the earth with a near circular trajectory relative to the earth. So add earth's orbital velocity (around the sun) to the moon's orbital velocity (around the earth). This will put the moon into an orbit around the earth, but might make it a bit more eccentric (elliptical). To correct this you can use angular velocity around the sun with ...

2

The energy is still there in the form of gravitational potential. Think of leaving the earth as a process similar to riding a bicycle up a hill. (When you ride a bike up a hill, you're moving against Earth's gravitational field thereby gaining potential energy, just like what happens to the rock when it moves up away from the surface of the earth.) If the ...

2

In the good old Newtonian world the gravitational acceleration is just: $$g = \frac{GM}{r^2}$$ The equation you give is just a rewriting of this. If you substitute: $$r_s = \frac{2GM}{c^2}$$ into: $$\frac { r^2 }{r_s} \frac {g}{c^2} = \frac {1}{2}$$ you'll find it simplifies to the first equation. So there is nothing especially meaningful in this ...

2

As far as I understand this problem, "GPE seems to exist only when I introduce the ball into the gravitational field" is a correct statement. As You said "You introduce the ball", so first You do the work that is converted to GPE and then GPE does work in accelerating the ball. I don't think there is a way in which a ball can appear somewhere within the ...

2

The two existing answers have both done the correct calculation, but both have forgotten to account for the Sun's gravity. A comet falling from the fringes of the Solar System is accelerated mainly by the Sun's gravity. We can see this from the expression for the potential energy at a distance $r$: $$V = -\frac{GMm}{r}$$ For the Sun $M = 1.9891 \times ... 2 When you say "without altering the actual momentum of it" is that really true? $$E^2 = p^2c^2 + m^2c^4$$ so for a photon$E = pc$, since rest mass is zero. Now according to your first "traditional" calculation of m, we would have$E = pc = m_1c^2$, and therefore$p=m_1c$, where$m_1$is mass according to the first "traditional" calculation. For your ... 2 Your equation is incorrect. The gravitational potential is $$\phi(r)=-GM\frac{3a^2-r^2}{2a^3}$$ when you're inside a uniform sphere of radius$a$with total mass$M$. This is a quadratic potential in$r$, which is why it gives rise to harmonic exchange of energy when you oscillate between the planet surface and the core. 1 I can't tell what your equation means, but from your words "which part of our planet matters more in gravity, the core, or what's around it", you can define $$X=\frac{M_c}{M}=0.29$$ where$M_c$is the mass of the core and$M$is the mass of Earth, as obtained from this link. Since gravity from a spherical body with varying radial density is only dependent ... 1 I am not sure what you meant by: "I figured I could simply calculate the magnitude of the components since that will give me the distance" But the idea is use the kinematics equations for x and y:$x(t)=x_{0}+v_{x0}t+1/2at^2$and$y(t)=y_{0}+v_{y0}t+1/2at^2$These equations are derived from integrating the acceleration function ... 1 A well executed barrel roll maintains the force balance you experience at rest with "gravity" oriented in the direction you experience as "down" (that is the direction from your head to your feet) due to centripetal acceleration. If you weren't looking outside, you might not realize the roll even took place (if the pilot is good). For those not convinced ... 1 Do you think anyone calculated the earth's speed to stay in orbit around the sun? As long as the speed is in the correct range the satellite will stay in orbit. For a satellite around the earth, the minimum speed is about 7 km/s. This is tangential speed, i.e. speed parallel to the earth's surface. Anything below 7 km/s, and the satellite will fall back. ... 1 You are correct - the force is constant in all four cases. Since each of the situations describes a "uniform spherical shell of matter," you can assume that the mass is concentrated at the center of that shell, as per the shell theorem cited. If you've learned Gauss's Law for electric fields, it can be applied to this problem. Gravitational force, following ... 1 Usually the Newtonian limit is described as taking$v << c$but a much better way to express it is saying that the kinetic energy is much less than the rest energy $$\frac{1}{2}m v^2 << m c^2$$ of course this runs into trouble when we talk about photons since we don't have a well defined concept of velocity, in the Newtonian sense. This is ... 1 For a particle of fixed mass$m$moving in a fixed gravitational potential$\phi(\vec{r})$the motion is independent of the mass of the particle. The equations are $$\vec{F}=-m\nabla\phi$$ and $$\vec{F} = \frac{d\vec{p}}{dt} = m \frac{d\vec{v}}{dt}$$ It's clear that the$m$'s cancel when combining these equations. So from this point of view it doesn't ... 1 The gravitational energy of the comet at infinity gets converted into kinetic energy of the comet. Calling$m$the mass of the comet,$M$the mass of the Earth,$r$the radius of the Earth we have: $$G\frac{mM}{r} = \frac{1}{2}m v^2$$ where$G$is the gravitation constant and$v$is the speed of the comet when it hits the surface. Thus: $$v = ... 1 Yes, you are correct in stating that the tension will be higher. In fact, it is simply:$$T=mg+ma$$It is important however, to make the distinction between tension and the maximum tensile strength. Tension, by definition is only as large as it needs to be (just like the normal force), because it is a reaction force. If it was any larger the body would ... 1 You may have noticed that if you start with the sun at rest, and put Jupiter into the system with an initial velocity to (say) the left, then over time the whole system moves left. (If you haven't noticed this is it worth setting the system up that way and letting it run long enough that you do notice it.) The trick is to recall that both bodies orbit their ... 1 This is the reduced mass. This effectively accounts for the fact that the center of the sun is not 'fixed', but rotates around the center of mass (also called the barycenter) of the Earth and the sun together. The expression$\frac{mM}{m+M}$approximates to$m$when$M>>m\$.

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