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12

First the somewhat misleading rubber sheet analogy: You've probably seen the bending of spacetime described as the deformation of a rubber sheet. Be careful taking this too literally as the sheet bending doesn't illustrate the bending of time, only space, and in any case it's not that good a mathematical model. Anyhow, it should be obvious that the ...


-1

No, every object's trajectory doesn't curve equally...for instance, light travels a null geodesic, which is very different from an orbiting planet. Your first example is quite correct, and the the path does depend on the momentum.But this applies to the second example as well...if the train is moving very fast, it won't bed at all. Since compared to gravity, ...


0

Consider a simpler case, the earth is only twice as massive as the ball, and suppose they are 3 meters apart. (And suppose they are tiny, so their radius does not matter.) They fall together and meet at their common center of mass, which is 1/3 of the way from earth to the ball. So, the work done on the earth is F * 1 meter, and the work done on the ball ...


1

Just to expand on vaaaaaal's answer, let's simplify this very slightly by assuming that the ball falls at it's average fall velocity $v$ for the whole height $h$ over a time $t$. Obviously, $v = \frac{h}{t}$. Then, we know that total momentum is conserved, so the Earth must fall up with speed $v_{e} = \frac{m}{M_{e}}v$. Thus, over the whole time of ...


1

The force from the earth on the ball and the force from the ball on the earth are in fact opposite and equal but the amount of work done on each is not the same. The earth is much more massive than the ball so, for an equivalent force, it is going to accelerate much more slowly and move a much shorter distance during the time the ball is falling than the ...


3

As Emilio points out the gravitational attraction by pluto is on the order of $10^{-14}\textrm{ m s}^{-2}$. However this is not an acceleration that can be felt in any meaningful way - In a flat gravitational field movement due to acceleration by gravity is identical to inertial movement. However as it happens the gravitational field from pluto is not ...


4

The distance at which the tidal forces froma primary start tearing apart a satellite is known as the Roche limit. In calculating the Roche limit we assume that the yield stress of the rock making up the planet is small compared to the gravitational forces at work so it can be ignored. The question is then simply whether the gravity of the body (in this case ...


1

For a book lying on a table, for example, the weight is cancelled by the upwards reaction force from the table. That's not quite true -- unless the table in a vacuum chamber at the south pole. The upward normal force exerted by the table and the downward gravitational force exerted by the Earth don't quite cancel. The book rotates with the Earth, and ...


0

I'm surprised none of the answers mention the phrase "normal force" which is what, I believe, you are asking about. JohnRennie's answer covers the concept, but doesn't use this term. Essentially, the normal force is the force exerted that resists gravity - you are pulled towards the earth by the force of gravity, the earth pushes back on you (keeping you ...


4

My knowledge is limited on the subject but matter is typically prevented from collapsing under the weight of extreme gravity by particle degeneracy. This is what keeps neutron stars from collapsing into black holes and is the result of particles resisting occupying the same quantum states. There are also some recent observations that indicate that there is ...


14

Suppose you're standing on a box as shown in (a) below: There are four forces acting. You apply a downward force $mg$ on the top of the box, and by Newton's third law the box applies an upwards force $-mg$ on you. The box transmits your force to the ground, so the box applies a downwards force $mg$ on the ground and the ground applies an upwards force ...


21

Yes, every gravitational force in Newtonian mechanics has an equal and opposing force, and it usually acts on other mass. More specifically, every two pairs of masses feel a gravitational force that's proportional to the product of their masses and inversely proportional to the square of their relative distance, but more important is the fact that both ...


2

I think you are doing your math incorrectly if you get $10^{21}kg$. $$M = \frac{Rv^2}{G}$$ Let's try Jupiter from your reference. $$M = \frac{(778 \times 10^6km) (13.1\frac{km}{s})^2}{G}$$ $$M = \frac{(7.78 \times 10^{11}m) (1.31 \times 10^4 \frac{m}{s})^2}{6.6743 \times 10^{-11} \frac{m^3}{kgs^2}}$$ $$M = \frac{1.34 \times 10^{20} \frac{m^3}{s^2}}{6.6743 ...


8

$$M_{gal}=Rv^2/G$$ ($G=6.67\times10^{-11} (N*[m/kg]^2) $. Units of $v$ and $R$ are km/sec. and km., respectively) You gave G in MKS, then: R and v are m, m/s, $ m= (\frac {1}{10^3}) km$, that's why you got a wrong result: $ 10^3 * (10^3) ^2 = 10^9 $ that's the order of magnitude you are missing $$ 1.5*10^{11} *(3*10^4)^2/(6.6*10^{-11}) = ...


0

My answer is the same as Chris', but formulated in a different way (it's essentially the same as this wiki article): In polar coordinates, the position vector is $$ \mathbf{r} = r\,(\cos\varphi,\sin\varphi) = r\,\mathbf{\hat{r}}, $$ with $\mathbf{\hat{r}}$ the radial unit vector. The velocity is then $$ \mathbf{v} = \dot{r}\,(\cos\varphi,\sin\varphi) + ...


2

The "associated scalar equation" is just the formula for the time evolution of the scalar magnitude of the displacement, $r$, rather than all its vector components. It really only makes sense to write such an equation if the right-hand side can be expressed in terms of $r$ only, and not $\mathbf{r}$. Then you can use it to analyze the evolution of $r$ in ...


1

Thermally supported, self-gravitating bodies (those to which the Virial theorem applies) qualify if you are willing to neglect the radiative energy loss. Depending on the time-scales that interest you this can be quite a good approximation. Stellar nebulae, brown dwarfs and so on.


2

The standard approach in numerical simulations is to do a discrete time stepping. If the motion is limited to the vertical direction, you just need to keep track of vertical position $z$ and vertical velocity $v$ (with positive $v$ denoting upward speeds). Starting with given values for $z$ and $v$ (the initial position and initial velocity) you update the ...


2

Assuming you are using C code: #include <stdio.h> int main(void) { float position = 0, velocity = 8, time, timeStep = 0.01; float g = 9.8; for(time = 0; velocity > -10; time+=timeStep) { velocity = velocity - timeStep * g; position = position + velocity * timeStep; printf("time %f; velocity %f; position %f\n", time, velocity, ...


0

Gravity is an acceleration. No negative involved. However, when you use acceleration to find a velocity, since velocity is a vector quantity, you must describe a direction. It is convention that anything that accelerates up, is described as a positive(+) like "The ball accelerates at 20m/s^2", whereas gravity describing a downward acceleration is described ...


-2

It is because gravitational force is attractive and work is done by gravitational force itself. When system does work itself energy is taken as negative and when work is done by external agency on system energy is take as positive.


3

You are not making a fundamental error, and your approach is in principle correct. Basically, what you get, is a discretization error. Basically, what you are doing is evaluating the integral $$d=\int_{t_0}^{t_1}v(t)dt=\int_{t_0}^{t_1} gt dt.$$ Which you approximated with a Riemann sum (maybe unintentionally?), i.e $$d=\sum_{i=1}^N g t_i \Delta t.$$ You ...


2

Your flaw lies in determining the number of meters the object falls in each second. Take just the first 1-second interval as an example. You stated it would fall 10 meters, while it will actually fall $gt^2/2=5\ \text{m}$. The basis of the flaw above is in your interpretation for $g$. This quantity tells you how much the velocity changes every second, not ...


3

g is telling you the constant rate at which the velocity is changing. So initially the velocity is 0 and after 1 second the velocity is 10 m/s. The average velocity during that first second is 5 m/sec so the mass has fallen 5 m. In the second second the initial velocity is 10 m/s and at the end of that second it is 20 m/s. The average velocity over that ...


2

Newton's 3rd law of motion is totally valid. Both the spheres exert on each other the same force which is $F = \dfrac{G(M_e)(M_s)}{(R_e + R_s)^2}$ . Their accelerations are only different but the force is same and opposite to each other. When u consider force of sphere on earth, u must consider the mass of earth and vice versa . I think u ve taken the $R$ ...


4

You equation for the surface gravity is only true at the surface of the sphere. Those extra 5 meters compared to the radius of the Earth are insignificant, however the other way around is far from true. So by using the sum of those two radii, when using one of the radii, will give you the right answer and if you look at the symbolic expression you should ...


0

If the force applied to each ball is the same then it will provide the same impuls to both, then using J = M(v-u) it can be seen that the lighter ball will reach a higher maximum speed. Following this, the ball's are now experiencing a drag force, depending on the assumptions made we can either assume that the drag force is constant, or that it is a ...


2

First one can get killed even by coming in contact (with speed from high altitude) with the water surface, which at this speed and momentum it appears as a "block of cement" (or more correctly, develop high enough forces to break your bones as per @dmckee's comment). This depends what wil be the impact surface (that is why seals and olympic divers fall into ...


0

This is probably a moot question anyway, since what objects are around it will greatly alter the formula. The moon, for example, has the gravity of the Earth to contend with which will cause more friction within the rock of the moon. If an object was orbiting Jupiter with all of its moons, it would have many gravitational factors which would not be present ...


0

The harmonic mean involves averaging the reciprocals, so it is dominated by the smaller of the two values. The reduced mass has the same requirement: The orbit of a small body around a large body is almost the same as the orbit of a small body about a fixed point. This also "explains" the factor of 2: in the limiting case m1>>m2, the reduced mass = m2, not ...


5

The book that you are following makes the following simplification: The object instantly comes to rest the moment it hits the ground. This is perhaps not completely physically reasonable, but I think you can also understand that the point of the relevant section is to discuss how far one can throw objects, not to provide a detailed discussion of the relevant ...


4

I'll take the question to be referring to solid rock. In reality, I think small asteroids are loose jumbles of rubble with a lot of vacuum between the rocks, and larger bodies like Ceres may have been liquid when they formed. Googling turned up [Scheuer 1981], which can be found online for free by googling. S/he estimates the maximum height of a mountain to ...


2

Some simple scaling relations suffice to determine the size beyond which gravity prevents non-spherical rocks from forming: A molecule of mass $m$ is bound to a mass $M$ of linear size $R$ with gravitational binding energy approximately equal to $G M m / R$. If this gravitational binding energy far exceeds the molecular binding energy $E_b$, gravity will ...


3

I am rather surprised that neither link posted above gives a simple discussion of the effect, so here it goes. Let us consider many asteroids of cubic shape, of constant density $\rho$, and of varying side $l$. We ask when, roughly, self-gravity will be able to perturb this shape into a spherical one. A cube of side $l$ has the same volume as a sphere of ...


0

There are problems mathematically with point particles, pointy surfaces, and the like. Point particles can be made to go zooming off with infinite velocity in finite time. Systems that violate the Lipschitz conditions can be set up, creating non-deterministic classical mechanics problems. These are conceptual problems. We use point particles because real ...


5

An interesting way to answer Part 2: Using the angular velocity version of the centripetal force equation:$$F_c=m\omega^2r=\frac{GMm}{r^2}$$If we assume that $r$ is both the orbital radius and the radius of the sphere being orbited and that the density of that sphere is $\rho$, then:$$M=\frac43\pi \rho r^3$$ then the equation ...


10

tl;dr: Velocity required: 1680 m/s Time to hit you: 6500 seconds Part 1: Velocity required (Using Google search values) Radius of moon = 1737.4 kilometers Mass of moon = 7.34767309E22 kilograms Assuming perfectly circular motion of the bullet, and no air resistance, and ignoring gravitational effects of other planets / objects in space, and using simple ...


0

There's an implicit question here, which is when does a conglomerate of particles count as a single object for the purposes of gravitation? The answer, in theory, is never. The earth isn't a single object; it's a conglomerate of many particles. When you're pulled by the earth's gravity, you're really being pulled by each particle individually. If you were ...


3

(Skip to the bottom for a list of classical and quantum-mechanical effects of gravitation that have been observed in subatomic particles; my attempt to explain quantitatively what it would take to measure atom-atom gravity got longer than I'd intended, and I haven't had time to shorten it yet.) Let's suppose you want to measure the gravitational attraction ...


15

Measure the gravitational attraction between two atoms? Heavens no. That's such a tiny, tiny attraction. The atoms will be attracted to themselves gravitationally, but only minutely. They'll be attracted gravitationally much more strongly to the Earth, to the lab setup and measuring equipment, to the buildings around the measuring equipment, and even to the ...


25

Groups in Seattle, Colorado, and perhaps others managed to measure and verify Newton's inverse-square law at submillimeter distances comparable to 0.1 millimeters, see e.g. Sub-millimeter tests of the gravitational inverse-square law: A search for "large" extra dimensions Motivated by higher-dimensional theories that predict new effects, we ...


3

This is one of those situations where you could argue on and on about definitions so let's answer it in all three meanings of the word weight that people might have. Does your mass $m$ increase when you inhale helium? When you fill your lungs with helium, you transfer $\approx1g$ of matter, which is several times less, but roughly of the same order of ...


0

Sun is greater 330,000 / 93 mil squared / 1(earth mass) x 250,000 mil sqared = ratio of g sun to g earth towards the moon at new moon. = 2.38 The ratio of sun to earth gravity is 2.38 , so the sun is more influential. However the moon is at a velocity that precludes it from fallimg into the sun. Just as the earth has has a velocity that allows it to ...


1

they will attract each other equally because there is no (known) thing that can cancel gravity. If they ever touch each other can't be told, because it depends on different things like the angle.


0

You know that SI units are arbitrary and not based on physical observations. I mean, kilogram, meter and second are arbitrary quantities based on what was reasonable in the 1700's. Their exact definition has been updated, but the amount their represent is arbitrary and bares no significance in nature. So there nothing special about any quantity defined in ...


0

Since air creates a force that is approximately proportional to the square of the velocity, the acceleration for each sphere is $a_r = kv^2/m (where \text{ } k = \frac{1}{2} C_x\rho\ S) $ The net acceleration on each sphere is $ a_n = g - a_r$. As the velocity increases, the $a_r $ increases until the net acceleration $a_n $ becomes zero $(a_r = g)$, and ...


1

Any inverse square law can be substituted by a Gauss law. In Gravitation, gravitational field $$E_{g}(R)=\frac{GM}{R^{2}}$$ Think of a sphere of Radius $R$ around the object of mass $M$ (This can be generalized to any shape). This gravitational flux coming out of it is $$\phi_{g}= E_{R}\times4\pi R^{2}=4\pi GM$$ So the gauss law will read as ...


5

That's a very hard question to answer theoretically because the aerodynamic drag, and therefore the terminal velocity, is highly dependant on the shape of the falling object. Assuming we're in the turbulent regime, the aerodynamic drag is given by the equation: $$ F_d = \tfrac{1}{2}\rho v^2 C_d A $$ where $\rho$ is the density of the air, $v$ is the ...


3

Like humans, the sheep has some – although more limited – freedom to act if it wants to change the asymptotic speed. A skydiver's asymptotic speed may be between 50 m/s in the "face down" free fall and 90 m/s when he or she pulls the limbs in. These two speeds hugely differ which indicates that it doesn't make much sense to talk about a "universal" value, ...


1

A black hole has two main features: a singularity an event horizon The event horizon is a sphere with a certain radius. Most people visualize the singularity as a point at the center of the sphere, and although that's not quite rigorously right, it's good enough for the purposes of the present discussion. Using a rough Newtonian analogy, the event ...



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