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21

Forget about force. Force is a bit much irrelevant here. The answer to this question lies in energy, thermodynamics, pressure, temperature, chemistry, and stellar physics. Potential energy and force go hand in hand. The gravitational force at some point inside the Earth is the rate at which gravitational potential energy changes with respect to distance. ...


11

There are two different quantities here to distinguish: the gravitational force and the gravitational well. At the center of the Earth, the gravitational force is zero, but the gravitational well is at its deepest. The heavy elements tend to migrate to the lowest point in the gravitational well, so they are at the center, even though the force is zero there. ...


8

Tall buildings are subject to the stack effect. Because the interior is held at a different temperature than outside (due to indoors temperature control), the vertical pressure gradient is different. Typically it is a reasonable assumption that the pressure inside the building is continuous from one floor to the next via the stairwell (if nothing else). ...


7

Why is the g=G∗M/R2 not working There are two regions to consider; the region outside the Earth and the region inside. The solution inside (non-zero mass density) and the solution outside (zero mass density) are different but must give the same value at the radius of the surface. Assuming a uniform mass density inside, the gravitational acceleration ...


7

I would say Brian Cox is being too cryptic. He is stating what is known as the Principle of Equivalence. In pure general relativity, gravity is not a force. It is the curvature of spacetime causing objects to obey the geodesic equation. This is a geometrical feature: the geodesic equation has no mass dependence. In free fall, the objects are unaware of their ...


7

Basically the rings don't fall into Saturn for the same reason the Moon doesn't fall into the earth. The rings are billions of little moons, each in it's own stable, or largely stable orbit. The rings are also likely resupplied with new ring material from Enseladus, Saturn's 2nd closest moon. (ice volcanoes due to strong tidal forces that can shoot ice ...


6

According to this lecture from the University of Edinburgh, numerical simulations of N-body systems suggest a half-mass relaxation time: $$ t_\text{rh} = 0.138\frac{N^{1/2}r_\text{h}^{3/2}}{m^{1/2}G^{1/2}\ln(\gamma N)} $$ where $r_\text{h}$ is the radius that initially contains half the mass of the system, $G$ is the gravitational constant, $m$ is the ...


4

While rare, there are a few uses of the method of images to gravitational problems. As lurscher says, the problem is finding equipotential surfaces. In most problems, such a surface doesn't exist, and hence the scare use of the method of images in GR. One class of problems for which it does applies are the so-called Dirichlet problems. Suppose one was ...


4

If you bring an object from infinity to a distance $R$ then the potential energy change is: $$ \Delta U = -\frac{GMm}{R} $$ Assuming your object starts at rest, the potential energy change is equal to the change in kinetic energy, so we have: $$ \frac{GMm}{R} = \tfrac{1}{2}mv^2 $$ so: $$ v^2 = \frac{2GM}{R} $$ You want $v \ge c$, so: $$ \frac{2GM}{R} ...


4

When you are lifting an object, you are exerting a force that balances the force of gravity on the object. By $$ F = m g$$ where g is the acceleration due to gravity, you see that a greater mass causes a greater gravitational force that has to be balanced by the force you apply to the object by holding it or lifting it at a constant velocity. Using the more ...


3

If the cable of the "elevator" is not connected to a point on earth, then the satellite must be in a geostationary orbit (or it will float away); this implies that if you now attach something to the platform (increasing the pull on the cable) you will pull the satellite down to earth. And as @lionelbrits pointed out, the pulling part of a space elevator ...


3

To expand on @Alfred's answer, The gravitational attraction from the Earth as you dig down into the Earth is due only to the mass of the Earth enclosed by the sphere defined by the radius that you are at (see the derivation of Gauss's theorem for a proof of this). Therefore, as you go further down into the Earth, the amount of mass that is pulling you ...


3

It is difficult to see how. Most comets and asteroids would encounter the Earth on a crossing orbit and the encounter velocity would be roughly the vector sum of the Earth's velocity around the Sun (of order 30 km/s) and the individual velocity of the rogue object. The individual velocity will vary from $\pm 30$ km/s for objects at a similar distance to the ...


3

Here's an interesting thought experiment. Imagine you have an elevator shaft to the centre of the Earth which, for some strange reason, doesn't affect the gravitational field of the Earth and doesn't flood with magma. OK, now at the Earth's surface get a bottle, half full with oil and half full with water. The water is denser than the oil, so the force ...


3

Even if I ignore wind and the drag forces and only consider the rotation of the earth the bullet will not hit the ground at the same place from where it was projected. There will be Coriolis effect. Coriolis effect: The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame. I suggest this ...


2

For two bodies this is relatively easy as the equations of motion describe a conic (an ellipse for a closed orbit, a hyperbola for an "open" orbit). You can use the vis viva equation to get the parameters of the orbit (semi major axis etc) from the given initial conditions, and the rest follows. For an ellipse, you can also express the position as a ...


2

As said by lemon, the reason is that the satellite drops to a lower orbit. This can be seen in a very simple way. Quantitatively The orbit can be approximated with a circular motion with a very slowly decaying radius. This implies we can write down equations for circular orbital motion: $$F = \frac{GMm}{r^2} = m\frac{v^2}{r} \Rightarrow$$ $$v = ...


1

It may be easier to consider a change in orbit caused by a pair of separate short thruster burns, to split the process into its various parts. Consider a satellite in a perfectly circular orbit around the Earth. Its speed is constant, say $v_1$. To go to a lower orbit, the satellite fires a thruster that opposes its motion. The satellite slows down, to a ...


1

I can integrate with respect to time, but that would give $$f(x,y)= \left\langle \frac{GMx}{(x^2+y^2)^{3/2}}t, \frac{GMy}{(x^2+y^2)^{3/2}}t\right\rangle$$ No it doesn't because both $x$ and $y$ are functions of time, so you cannot use this "simple integration" method. What you need to do is go back to the differential equation, $$ m\ddot{\mathbf ...


1

but if you can give me a hint that would be great. Since the force depends on the radial distance only and points towards the origin, the angular momentum (assumed to point along the $z$ axis) is conserved. This suggests that the appropriate coordinates are the spherical polar coordinates $(r, \phi)$ where $$x = r \cos \phi$$ $$y = r \sin \phi$$ In ...


1

The thing about Lagrange points is there is always two ways of viewing the system. Method 1: The Inertial Frame. Position yourself "above" the Solar System, looking down at the objects moving in ellipses. Here the force of gravity is the only force. Earth doesn't crash into the Sun because the constant inward acceleration merely bends its path into an ...


1

On a practical level (baseball, soccer, tennis, ...) Newton's account of gravity as a force with particular behaviors is entirely satisfactory. It is even used for launching satellites (well most of them), and predicting artillery impact points. It was this last application that made warfare even more horrendous than it had been (and safer for the ...


1

An object at L3 orbits the Sun through exactly the same mechanisms that the Earth does: it feels the Sun's gravitational attraction, and this is exactly the centripetal force it needs to perform a uniform circular motion about the Sun. The slight problem, as you point out, is that it also needs to contend with the Earth's gravity, which at L3 also pulls it ...


1

The answer was in fact covered by the Curious Mind, but for you to see the process how the potential energy transforms into kinetic, here is an elementary elaboration. The equation of motion in the gravitation field says that $ \ (1) \ h_0 - h = \int _0^t v(t) \ \text d t $ Multiplying this equation by $mg$ which is constant $ \ (2) \ E_P(0) - E_{P}(t) = ...


1

In general relativity, the gravitationally free-falling objects are inertial, while you standing on the Earth's surface are accelerated upwards by the force provided by the floor you're standing on. That is why you see the objects as accelerating downwards--because you are in an accelerated frame. Therefore, your understanding of mechanics on this probably ...


1

The method of images works on the electrostatic case because the axis of symmetry of the mirror charges induces an equipotential line that is equivalent to the infinite conductor surface. In gravitational physics, there are no known instances of a physical surface that is at the same potential in the gravitational field.


1

I think you could possibly engineer such a collision between two bodies in horseshoe orbits. The minimum mutual speed between the two bodies depends on their mass difference, with a limit that approaches zero speed if the bodies are the same mass. I haven't tried very hard to do this, but it's the only possible way around the escape-velocity argument put ...


1

It would help me, but even more importantly you, if you defined what your symbols are supposed to mean. You did that with $U_i$ and $U_j$. So let's be precise together, as an exercise: Let's explicitly state that we consider a point mass $m$ in a gravity field caused by a much larger mass $M_E$ at the origin of the coordinate system, so that we can assume ...


1

$U$ in your equation in potential energy, and $W$ is internal work. That is, the work done by forces within the system. The system in question here comprises the object and the earth, and the internal force is gravity. The work that you do to lift an object is external to the system, and does not appear in your formula. In your scenario, lifting an ...


1

Let me assume that the object has spherical symmetry, however, for solving the present problem it is painted on its surface with different colors. So, imagining a plane section that contains the orbit of the object around the earth. The section of the plane through the object is a circle and we will see different points of the circumference painted in ...



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