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8

Visualization The difference in height $h$ is always the same (here 10 m)! Remark This is of course only true if $g$ is constant, e. g. $h$ does not change "much". See also Wikipedia.


7

Your misconception has nothing to do with gravity - you're just getting a little mixed up about acceleration vs. relative acceleration. Let's dispense with gravity, since it's a red herring here. Say there are two cars. Car A accelerates at $+3 ~\rm m/s/s$ (to the right). Car B accelerates at $-5 ~\rm m/s/s$ (that is, to the left). So far, so good, right? ...


5

How “large” is a Lagrange point? L1, L2 and L3 are essentially zero size cause they're never stable. They're still useful cause an orbital near L1, L2 or L3 doesn't require a lot of energy to stay in that general area. so we can use L1, 2 or 3 for not quite stable orbits that don't require much energy adjustment. L1 Halo orbits are used too, not ...


5

In classical gravity, the answer is "never". In general relativity, the answer is "never". Now what about a quantum theory of gravity? We don't know how it'll work, but it should reduce to general relativity in the classical limit (i.e. the limit of weak fields and large distances, which is exactly what your question is about). So the answer is still ...


5

Potential energy is given only as a difference of energies at different heights. So, if you want to know just how much does the person gain energy (or rather loose by friction in their muscles and joints) by walking down the hill, you might just use their height of their heels on the top of the hill and under the hill. But remember, you always have to use ...


4

I'm going to answer the question of your title, and also address the curious statement that "tidal gravity=real gravity". Let's begin with your statement: Tidal gravity, by my understanding, is the difference in gravity between two points. You're very much on the right track here. When people talk of "tidal effects" and "tidal gravity" when not ...


4

Because in the second scenario you described, the exhaust gases from the rocket fly off at high speed into space. This is where the extra energy goes. A better alternative to get circular motion of the satellite with the earth removed, would be a second satellite, connected to the first satellite with a long cable. If, for example, both satelittes have the ...


4

The answer is basically, angular momentum. The collapsing proto-solar nebula has some angular momentum. Whilst dissipative processes can allow the nebula to collapse along the axis of rotation, there is still the problem of how to shed angular momentum in order to allow gas/dust to orbit closer to the rotation axis. This is just a basic application of ...


4

You need to be careful with the units. On the Wikipedia page which gives the gravitational parameter of the Earth as $398,600$, the units are km$^3$s$^{-2}$, so this is $3.986E14$ m$^3$s$^{-2}$ in agreement with your figures for $G$ and the mass of the earth in SI units.


4

Theoretically, their center of mass is what you're looking for. It's somewhere near the stomach. High jumpers bend their body when they are jumping so that their center of mass can travel just above the bar which allows them to use the least energy to jump the highest: https://en.wikipedia.org/wiki/High_jump When a body is rotating or something unusual is ...


3

First off, a definition of gravitational acceleration from the perspective of a geologist or geophysicist. Imagine a 50 cm tall cylinder with all the air drawn out. A small ball is held in place at the top of the cylinder. The device is firmly fixed to the surface of the Earth and the ball is released. Timing how long it takes for the ball to fall to the ...


3

Dimensional analysis should never take the place of common sense - it should be used to inform it. Let's take Newton's example. He knew that objects of different mass fell at the same speed - which had to mean they experienced a gravitational force proportional to their mass. Hence $$F \propto m$$ If gravitational attraction between two bodies is ...


3

No distance is far enough. Among other things, if you are extremely far away, then there is room for lots and lots of things to be far away from you and even if they individually have little effect we can find the net effect of all of them. So we know the effect of each one is not zero. So we can prove the effect of A on B is not zero even when they are ...


3

If they're at rest to start with, and if there is nothing else in the Universe, then Yes. The gravitational force increases their velocity to a finite value (which still increases) after a finite time which is enough for them to collide at some moment. The time may be very very long if the objects are very light and/or very distant, however. (You may avoid ...


3

Our mass is so small that large, I mean large enough to be visible things, like a fly, could (and do) easily escape our bodies tiny gravitational pull, but smaller things, like an atom, may for a short while be stuck in orbit around us, but we can't detect them very easily, if at all. Gravity is a very,very,very weak force. A related answer Humans And ...


2

If someone (like superman) could stop the moon from its orbital motion then yes it would fall towards the Earth. Only then the direction of motion would be parallel to gravity. Same with the I.S.S or the satellites orbiting Earth. They could also spiral in and crash because the atmosphere is taking away their kinetic energy. Just like the comment says ...


2

According to Google the mass of the earth is around $5.972 \times 10^{24}$ kg (presumably that figure includes its 7 billion human inhabitants plus everything else were meant to be sharing with). If we take the average weight for a human as given by WolframAlpha as 62 kg this gives humanity a combined weight of $4.34 \times 10^{11}$ kg - so our share in the ...


2

General relativity in just one dimension will always be flat, as all 1D manifolds are diffeomorphic to flat space : \begin{equation} ds^2 = -f(t) dt^2 \end{equation} As you can perform the variable change \begin{equation} \frac{dt'}{dt} = \frac{1}{\sqrt{f(t)}} \rightarrow t' = \int \sqrt{f(t)} dt \end{equation} Giving you \begin{equation} ds^2 = - ...


2

In my experience, in brushes the strongest force acting on the water is the capillary force due to the surface tension in the liquid and the proximity of the hairs in the brush. Surface tension will cause the water to try to "wet" as much of the brush hair as possible - regardless of orientation. If there is excess water, such water will be pulled down by ...


2

I assume that you want both particles to be free to move, each in a circle. That circle should be about the center of mass of the system, which will be at $$\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}.$$ The radius, $R$ of each circle will be the distance from the mass to the center-of-mass point. Calculate the Newtonian gravity force on a particle, set it ...


2

For common center (barycenter) orbits, the velocities will be \begin{align} v_1&=\sqrt{\frac{Gm_2r_2}{\left(r_1+r_2\right)^2}}\\ v_2&=\sqrt{\frac{Gm_1r_1}{\left(r_1+r_2\right)^2}} \end{align} which, since $m_1=m_2$ and $r_1=r_2$, will be the same value, $v\approx0.22$ for your values of $G,\,m,\,r$. Since you've placed the two objects along the $x$ ...


2

Your question is about finding the initial conditions to a circular orbit, which has already been answered very well. This answer instead tackles determining the initial conditions for more general orbits. This will allow you to simulate different orbit shapes. Below are a few shapes of different orbits: There are two parameters that govern the shape: ...


2

W=mg for the motorcycle rider and the person in the airplane, whether they are in free fall or not, because, as you stated, the mass of the person is unchanged and the earth still exerts gravity on that person, during their free fall experience. When you get back to the ground, or the airplane gets back to level flight, the surface that you are standing on ...


2

It doesn't have to be modified, it's fine as it is. There's no paradox at all. The force that attracts both to each other is indeed $F=G\frac{M_A M_B}{r^2}$. But the acceleration they experience is not the same (at least provided $M_A \neq M_B$) because their inertias (masses) are not the same. For one $F=M_A a_A$, for the other $F=M_B a_B$. There's no ...


2

The gravity g from a body of mass m at d distance from the center of the body can be found by the equation g = (m × G) / (d2), where G is the universal gravitational constant. Usually, g is in meters per second2, m is in kilograms, d is in meters, and G is in meters3 per kilogram per second2. If you want to apply this to the Earth, the Earth's mass is ...


2

There is a confusion of what that $v$ means. You are thinking about the velocity of the drone, which is stationary this $v_{drone}=0$. But in your equation, to calculate the power needed by the wings, you have to consider the velocity of the motor providing the thrust (propelling air downwards at a certain rate) to keep the drone floating which is making ...


2

I will simplify this problem by assuming that the only forces come from Newtonian gravity and by limiting the masses of the moons to much smaller values than that of the planets, such that the moons exert a much smaller gravitational forces on all other bodies and therefore can be neglected; so the only sources of gravitational forces are the two planets. ...


1

If you have a spherical body of radius $R$ with mass $M$, the gravitational field at any point at a radial distance $r$ is given by: $$\phi=\frac{GM(r)}{r^2}$$ where $M(r)$ is the mass enclosed inside a spherical shell of radius $r$. This is the only mass that matters in this case. (because of the Shell Theorem: https://en.wikipedia.org/wiki/Shell_theorem) ...


1

Limitations of $g \propto \frac{1}{r^2}$: The relationship $g(r) = G \frac{M}{r^2} \rightarrow g \propto \frac{1}{r^2}$ (where $g$ is the acceleration due to gravity, $G$ is the universal gravitational constant, and $r$ is the distance between the massive object and the accelerating object) is just fine in Newtonian physics—it doesn't need to be fixed. ...


1

If this is a correct description of what happens, can we conclude that g does same work on P and on P'? Yes. This is correct. If g acts perpendicularly to the velocity, it performs work of magnitude zero. This is also correct. The reason the two statements above are not contradictory is that the work done by the gravity changes the direction of ...



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