# Tag Info

## New answers tagged newtonian-gravity

0

Ultimately, a mystery of sorts, but see https://en.wikipedia.org/wiki/Dirac_large_numbers_hypothesis and google "Dirac large number hypothesis" for what might (or might not) be a step towards explanation. For example, the relation $GM/Rc^2\approx1$, where $M\sim\mbox{"mass of universe"}$ and $R\approx13.7\times10^9\mbox{l.y.}\sim\mbox{"size of universe"}$, ...

0

Newton just find that the gravitational force is proportional to the product of mass and the inverse of the square of the distance. Some years after, Cavendish mesure the value of that constant G. That constan could be 1, or 1000000. But G=6.674×10−11 N · (m/kg)2. Why?, nature works that way.

-1

like the quantum mechanics if you could not measure something so that not exists/ how could you detect a mass if there is no force even if you get close to that to touch it you have a mass yourself so the force would appear and if not you are contradicting newtons law of gravity so the argument is not that simple mass and force are defined by each ...

0

Your video is appropriate, the sun is in place and all the rest, including the fascinating scenery. Remember, if you're thinking of adding anything else, do not move the object hitting the sun, whatsoever. The sun's gravity is unimaginably powerful, so making bouncing effects would seem unrealistic. Also, as the sun has strong heat, (I know you mentioned ...

0

I can provide only "classically newtonian" answer cause my orientation in general relativity is only very limited. As it's written in the wikipedia article, the $G$ is empirical constant. How to understand that? Experimental scientists are usually aware of dependency on parameters, but the absolute values are corrected using various additive or (in this ...

2

This video by Richard P. Feynman might explain how hard it is to answer such a 'why' question. An excerpt: But the problem, you see, when you ask why something happens, how does a person answer why something happens? ... When you explain a why, you have to be in some framework that you allow something to be true. You have to know what it is that you’re ...

3

Yes, Newtonian Physics works on a galactic scale. Still, for long distance interactions on fast objects you might want to take into account the finite speed of gravity, but I don't think it is necessary for ordinary galaxies simulations. Conversly a lot of phenomena occur that impact the galactic material: writting a decent simulation is not easy.

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There is a mutual attraction from gravity, and we generally only consider the smaller object here on earth because the earth is so massive, the acceleration of the earth is negligible. This is because $a = F/m$, and with equal $F$ between the two objects, the acceleration will scale as $a\propto 1/m$. For the earth, this leaves $a$ ridiculously small, but ...

2

Does the bigger mass EVER move towards the smaller mass? Yes. $F = KMm/r^2$ $M*a_{M}=F$ $m*a_{m}=F$ As you see the smaller the mass the higher the acceleration and in consequence the higher the traveled distance in a given time t. If the above is true, can we technically move the Earth by us(human population) jumping indefinitely? No. Each ...

1

yes, the earth will accelerate towards you , however the Earth's acceleration will be so small for all practical purposes that you usually do not consider it. Earth's acceleration is small because the mutual forces between you and the earth are the same, but the masses are different, so this results in different accelerations (remember: $F=ma$). Now if you ...

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In all cases, the two objects move towards one another. In fact they experience exactly the same gravitational force. However, because acceleration equals force over mass $$\mathbf{a} = \frac{\mathbf{F}}{m}$$ that equal forces causes the heavier object to accelerate much less than the lighter one. But technically, the Earth does move towards you very ...

2

Well that depends. As always, you should be very careful with such reasoning as Due to the fact that gravity is related to the square of the distance should not the gravitational sum of every particle exceed the force when calculated by the center of mass. because this is a problematic statement. In general your (Newtonian) gravitational potential is ...

1

Strictly speaking, $g$ (even more strictly speaking, $g_0$ or $g_n$) is a constant. It is exactly 9.80665 m/s$^2$, by definition. There are many places in science and engineering where it is very useful to have an exact (albeit arbitrary) defined constant for gravitation on the surface of the Earth. That said, gravitation on the surface of the Earth does ...

4

To two significant figures, the acceleration due to gravity is $g=9.8\:\mathrm{m/s^2}$ everywhere on Earth (at sea level). That is to say, if you use e.g. a pendulum to measure $g$ to two sig figs, you will get this value no matter where you are. In a sense, this is the precision to which the Earth is well-approximated by a uniform sphere of matter. The ...

0

You can also read about the test of 'frame dragging', which is predicted by General Relativity and not Newtonian gravity, confirmed by Gravity Probe B. Also, Newton's law of gravity is not so much 'wrong' as it is 'incomplete' (General Relativity reduces to Newtons law in the case of low energy/speeds).

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In addition, one more test will be the gravitational wave from binary, supernovae as well as supermass black hole.

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There are three classic tests of general relativity: the anomalous precession of Mercury's orbit the deflection of light by the Sun the gravitational redshift of light Newton's theory predicts zero precession in test (1) and zero redshift in test (3). For test (2) Newton's theory predicts a deflection half the size of the prediction in general ...

4

Actually you can go to the orbit of Jupiter with a $\approx 2500$ tonnes rocket and a $3$ tonnes payload. From there you can use an ionic engine. A rocket launched from the equator of Jupiter that turns at $12.6~\text{km}/\text{s}$ needs just an increase in speed $v = 29.5~\text{km}/{\text{s}}$. $$v_{rj}:= 12.6~{\text{km}}/{\text{s}} \;\;\; R_j := ... 16 Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then: radius of Earth = 6.4 \times 10^6~\text{m} mass of Jupiter = 1.9 \times 10^{27}~\text{kg} Escape velocity, v_\text{escape} = \sqrt{\frac{2GM}{r}} This gives a value for ... 10 Hey! The question keeps getting edited! Make up your mind! You asked about Mars originally, then edited the question. Actual, real Jupiter is flat out impossible. Does it have a surface to launch from? Who knows? What's the pressure at that depth? Can our probes even survive at that depth? Probably not? What if Earth had the mass of Jupiter? More ... 1 If the planet is spherically symmetric, the gravitational field is the same as if all the mass were concentrated at the center, so the attraction will be along the line toward the center. If it is not, it is not really fair to talk of the center. Imagine if there were a huge mountain at point A. This could change the gravitational field (a small bit) to ... 2 The moon has a large tangential velocity. If you throw the apple parallel to the ground, it also moves in a curved path. 2 As @tmwilson26, said the net force is directed along the center, here's the mathematics of it if there is a small differential mass at A, there also exists another point A' laying just opposite to the center line(making a same angle \theta with the center line) The force F experienced at point O has 2 components F_x and F_y where where F_y is ... 9 You have to think of the symmetry of the situation. For each point A and B, there is a point A' and B' opposite the center line that has an attractive force pulling in such a way that the components perpendicular to the center line are canceled. Therefore, the net force is along the center line. 0 No, unless you shoot the object downwards with a cannon providing it with an initial velocity higher than the terminal one. Regarding your example, sky divers do decelerate. In groups they use different (maybe unstable) positions in order to change their resistance to air and thus reaching speeds higher that the terminal one (in some stable position, like ... 2 No. the earth do not accelerate with 9.8 meter per sec. squared. The acceleration is quite negligible. And as the objects on the whole earth is distributed almost uniformly the net acceleration due to the objects is zero. That's why we do not notice the earth moving around like that. MATHEMATICALLY we know that | F_E | = |F_o| where F_E is force ... 0 As far as we know, everything that has mass has "gravity." If you drop a grain of sand, the reason why the grain of sand falls to the ground is not just because the Earth attracts the grain: It's because the Earth and the sand grain attract each other. The force that attracts one massive body to another is given by Newton's Universal Law of Gravitation: ... 1 The acceleration of gravity is the weight of the object on the surface. If the structure, either of the object or the surface, is deformed by the weight, then part of the gravitational potential energy of the object will turn into kinetic energy of surface moleucles and therefore heat, and this will be shared between the object and the surface. In this ... 0 With respect to current theory my answer may prove quite unpopular (down votes expected), so please take this answer with a grain of salt if you are still in school. Your approach in this 'thought experiment' that you propose is very much like some of my own research conclusions. In direct answer to your question, yes, General Relativity should require an ... 0 In general, the position is the integral of the velocity - this is the essence of the equation you wrote down (without the "terminal velocity" part which makes no sense). If, at a certain time, you change the velocity (direction), you can just treat that point as the start of a new trajectory: the initial velocity will be "the velocity of the turned ... -1 A body at rest in a gravitational field ((as standing on the surface of the earth) is the same as a body being accelerated. A free falling object is equivalent to not accelerating (The equivalency principal). The Accelerated one at rest would heat up. OK so maybe I didn't articulate that so well please let me try again. The equivalence principle states that ... 2 If the object is at rest it would imply that gravity transfer heat without an increase in potential energy, and there are no other forces that produce work. This would violate the conservation of energy. Regarding your comment "any contact forces imposed on an object will increase that object's heat energy": this is incorrect, friction only results in heat ... 1 God No. Here is why: Deceleration is a decrease in velocity, i.e dv/dt is negative. With terminal velocity,it just becomes zero. You can see the same thing if you plot a graph. For the skydiver's Velocity -time graph, the slope of the curve( acceleration) is NEVER negative. It becomes zero at terminal velocity. 2 Does an object decelerate when reaching terminal velocity No, it ceases to accellerate. So while the force of air resistance increases to equal the force of gravity is the sky diver decelerating? Consider F = ma. You say because F_r = -F_g, net F is zero. What does that imply about a? 2 What is gravitational potential? Usually a potential is defined as the potential energy per mass or per charge or similar. This is most often seen in relation to electricity or chemistry and less often to gravity. GPE is gravitational potential energy. GP is gravitational potential energy per mass:$$GP =\frac{GPE}{m} $$Is it defined for the system ... 0 First ask yourself what Newtonian gravity would say. Would a particle orbit around the center of a perfectly spherical star, or would the star and particle co orbit their common center of mass? Obviously the latter, but the former is a good approximation where you ignore the effect of the particle on the star. So how about general relativity? Same deal. If ... -4 I was reading about Supplee's paradox, which is about whether a relativistic projectile, subject to uniform gravitational acceleration, would float or sink underwater. However the solution of the paradox seems a little unclear to me, as given in the link. It isn't very clear, is it? Given certain assumptions about how to treat the gravitational ... 1 Your copy of Verma has already defined gravitational potential energy previously in (11.3) The gravitational potential energy of a two particle system is$$U(r) = -\frac{Gm_1m_2}{r} \tag{11.3}$$where m_1 and m_2 are the masses of the particles, r is the separation between the particles and the potential energy is chosen to be zero when the ... 0 This is the elliptic case of the radial Kepler problem, the equation for time as a function of position is$$ t(r) = \sqrt{ \frac{d^3}{2 g} } \left( \arccos\left( \sqrt{ \frac{r}{d} } \right) + \sqrt{ \frac{r}{d} \left(1 - \frac{r}{d} \right) } \right) $$where t is the time, r is the position, d is the initial (maximum) separation, and g=G(m1+m2). In ... 2 The spacetime outside a spherically symmetric arrangement of mass is described by the Schwarzschild metric. This is a consequence of Birkhoff's theorem. So the changes in the interior structure of your object make absolutely no difference to an object outside it. The orbit will be exactly the same as if the object was unchanging, or indeed if it was a black ... 0 Actually it is similar to the solution of a body going through the center of the earth through a drilled hole . Neutrinos could go through such an orbit with zero angular momentum as the probability of interacting and disappearing is very small. Planets cannot. If Goldstein is in a chapter for planetary orbits (hint Keplerian) it is obvious why the one ... 0 The two equations of motion reduces down to one equation of motion by considering the separation x=x_2-x_1 and the separating acceleration \ddot{x} = \ddot{x}_2 -\ddot{x}_1$$ \ddot{x} = -\frac{G (m_1+m_2)}{x^2} $$or  \ddot{x} = -K/x^2  with K=G (m_1+m_2) This can be re-written as \frac{{\rm d} \dot{x}}{{\rm d} t} =\frac{{\rm d} \dot{x}}{{\rm d} ... 0 What you have is a system of coupled differential equations. Say the position of the masses are m_1 and m_2. The positions are x_1(t) and x_2(t). We'll assume that x_1<x_2. Note that they will stay on a line, so it suffices to consider one dimension. Now, we use F=mx'' to construct our ODEs:$$G \frac{m_1 m_2}{(x_2(t)-x_1(t))^2} = m_1 ...

1

I assume that friction is an external velocity dependent force in your simulation code. Since you have such external forces, your total energy, total angular momentum, total momentum are likely not to be conserved. In your case, the friction is a phenomenological external force, but similar behavior could also be simulated with a large particle, moving in a ...

1

Your teacher is wrong. The gravitational potential $V(x)$ is generally defined as potential energy per unit mass i.e. $V(x) \equiv \dfrac{U(x)}{m}$. So for the points where $U(x)$ is zero, $V(x)$ is zero and vice-versa by definition. EDIT: After you added the comment and a snapshot of the book, I realized your book has defined Gravitational Potential in a ...

3

According to conservation of momentum, the center of mass of a system cannot accelerate without external forces. In other words, if the center of mass starts out at rest (which is generally a good procedure in simulations), then it should always stay at rest. It is normal for numerical errors to introduce deviations, but the motion you are seeing looks ...

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