# Tag Info

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I don't want to prove you that mass generates gravity. As ja72 already said there are hundreds of years of experience on that. Instead I just want to show you how weak and refutable are your arguments. My opinion is that it is the rotation that causes gravity, when we see a whirlpool in water it takes anything that comes near to it in to the center ...

3

See moment of inertia is analogous to mass. Moment of inertia can be thought of as a physical "property" of the object similar to that of mass. And as we know that mass does not depend on any force or gravitational field or any other external effect, so does moment of inertia. Hope this answers your question.

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Your mistake is in your conservation energy equation. The way you wrote it is valid only when falling from infinity, from rest. The correct is: $$dE=dK+dU=0,$$ that is $$mvdv=-\frac{K}{x^2}dx,$$ where $K\equiv Gm_1m_2$. Integrating from $(x_i,v_i)$ to $(x,v)$ we get $$\frac 12m(v^2-v_i^2)=K\left(\frac{1}{x}-\frac{1}{x_i} \right).$$ This is the correct ...

1

In addition to the already given answears this also might be of interest: When hammer and feather are dropped simultaneously they arrive at the same time, when dropped independently the hammer attracts the planet more than the feather, so you are right, the total time until impact is then smaller for the hammer. If you pick up the hammer and let it fall to ...

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The point is if all effect was taken into account. Math would be summed up that effect of more mass under your feet still less than effect of distance from the center of mass Another view is. At equator there are bulge near you. But from all other side of earth the bulge is far from you. Compare to the pole that all bulge is equally far from you, that ...

2

The gravitational field of a spherical mass is the same as the gravitational field of a point mass at the centre of the sphere. This result is known as the shell theorem, and it is why we measure distances to the centres of spherical objects like planets or stars. However when the object is not a sphere then as a general rule its gravitational field will ...

-1

No, we don't know that for sure about gravity. One object cannot have 2 different masses: The force that makes things roll, and fall, is the same force; it is gravity. This is what my question is about. We don't know for sure that Newton was correct: The fact that something heavier rolls at a higher speed shows us that Newton's theory of gravity is wrong. ...

1

It's called the virial theorem: in a bound system, the average of the potential energy of a $\frac{1}{r}$-potential V is related to the average of the kinetic energy T like $\frac{\langle V \rangle}{2} = - \langle T\rangle$.

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The tangential velocity is related to the radius of a circular orbit and the angular velocity of the orbiting object. Specifically: $$\overrightarrow{v}=\overrightarrow{\omega}\times\overrightarrow{r}\to v=\omega r$$ For the satellite, the centripetal acceleration is equal to $r\omega^2$, which is the same (according to the above relation) as the tangential ...

1

The relation is indeed correct. Let me show you why. This type of problem is solved by equating forces. Other classical mechanics problems are by equating energies or, in rare cases, momentum. The gravitational potential indeed is $$V(r) = G \frac Mr \,.$$ This is not directly useful for this problem. Here you rather want to look at the gravitational ...

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John above spoke about the law of equivalence to refute your idea, however, a new theory by a British university teacher claims that there is a difference between "an elephant and a gnat" falling down (John's example). MIT wrote about it the other day too: ...

2

You ask: could the gravitational force of an average asteroid be extrapolated to the rest of the asteroids in its family? But the question is what properties the members of a family have in common. You could argue that all the asteroids in a family will be made up from the same material and will have the same density. That means as long as you can ...

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The cross product of two vectors is the area of a parallelogram defined by those vectors. If the angle is 90 degrees then the cross product is simply the magnitude of vector 1 multiplied by the magnitude of vector 2 (i.e are of a rectangle). another way of writing the cross product is |v||u|sin(theta). If theta is 0 (both pointing in same direction) or 180 ...

0

Let $x = x_2-x_1$ and $y= y_2-y_1$ to simplify the algebra. The initial velocity $v$ has two components $v_x$ and $v_y$ and the angle of projection is $\theta$. For vertical motion you have $y = v_y t - \frac 1 2 g t^2$ and for horizontal motion you have $x = v_xt$ Combining these two equations gives you $y = v_y\left (\dfrac {x}{v_x} \right) - \dfrac ... 0 If you ignore air resistance, it can be solved as follows. Assume the initial velocity at$t = 0$at$(x_1,y_1)$be$(v_{x0},v_{y0})$. Assume it hits$(x_2,y_2)$at$t=T$. Solve for$(v_{x0}, v_{y0})$in terms of$T$. You then get a family of curves when substituting different$T$. 0 I will give you a hint: We can avoid the constant c in the equation by selecting the starting point of parabola as origin. so we have three points$(0,0)(x_1,y_1)(x_2,y_2)$. Now placing it in equation$y=ax^2 + bx$. We can find a and b. Now comparing with the equation of trajectories we can find out the velocity$v_0$. and angle$\theta$in form of either ... 0 For orbit, try thinking of it this way... First simplify from three dimensional space to a 2 dimensional plane, say a sheet of paper. Also for simplification, we will consider the case of one large object and one small object. We will consider the large object to be 1) not moving and 2) so much larger than the small object that the large object is ... 1 The classic thought experiment (which I think is in Feynman's lectures) is to imagine a cannon on top of a hill shooting cannon balls. Ignore air friction for this. Typically the ball curves in an ellipse (we often say parabola, but this is an approximation for a uniform gravitational field, though the two are basically identical for this case) and hits ... 0 While the motion of airplanes is very different from the motion of moons, planets etc. motion is the reason aircraft stay aloft (with the exception of lighter than air cart - hot air balloons etc). To counter gravitational pull plane's wings generate lift by their motion through the air. If an airplane moves through the air too slowly it will stall and ... 0 Planets (take for example the solar system) revolve around a huge massive star under the influence of gravity exerted by the massive object on the planet if the planet is under the field of influence of the star. They are actually falling towards the star, but each time missing it from colliding. Consider the moon. It is constantly falling to earth under ... 1 See this Figure For point mass$m_1=1$, and box mass$m_2=3$, and box size$a$. 0 If you consider the case of a point charge (that we can take as positive) above a grounded conducting plane (not quite your problem) you can use the method of images to determine the electric field. In this case the field lines with be perpendicular at the surface of the plane, and this results in an induced surface charge which will be negative. I do not ... 2 A helium nucleus is not two protons, it is two protons and two neutrons. Instead you can compare two deuterium nuclei (one proton and one neutron) with one helium nucleus, so you have the same number of nucleons of the same types. And the answer is (as far as I am aware): no, there is no experimental evidence for the magnitude gravitational forces produced ... 5 TL;DR: Yes. Although in reality you don't have unlimited floating point precision, and this will almost always break time-reversibility. I should point out that not all integrators are time-reversible. For example, predictor-corrector schemes, and most schemes that deal with constraints. The Verlet method, however, is time-reversible, even for large ... 3 For your second case, you can change the angular momentum, but remember that you have fixed your total energy. You can't make the planet revolve arbitrarily fast or it will have more energy than allowed. By increasing the angular momentum without adding energy, you are circularizing the orbit. To add, you might take a look at the Specific Orbital Energy ... 0 Whenever a pendulum moves, it accelerates since the velocity vector is constantly changing. So any pendulum that 'pendulates' forever will 'accelerate' forever. However, I think the acceleration you are talking about is better described as angular acceleration. Perhaps a better way to describe it is if it will complete a full rotation with increasing ... 0 If the pivot is accelerating horizontally (together with the body) at a rate of$a_{pivot}$then the angular acceleration of the pendulum is $$\ddot{\theta} = - \frac{m c (a_{pivot} \cos\theta + g \sin \theta)}{I_{zz} + m c^2}$$ where$c$is the distance from the pivot to the center of mass,$m$the total swinging mass and$I_{zz}$the mass moment of ... 1 @ Nick Basically you are confusing two different coordinate systems. In the coordinate system where you decomposed g into two components, there the question of x and y doesn't seem valid because you have transformed the cartesian coordinate system into a different one. You would have new x and new y there and you have to stick with one while solving the ... -4 I was considering this question as well. Neither matter/energy can be created or destroyed, only converted from one form to the other. Consumption of star stuff by a singularity would convert that star stuff into star energy, but with no other way of expressing that energy there seems no choice but for nature to convert with 100% efficiency that matter into ... 1 John Rennie has explained the problems with units here. Now, you'll burn about a factor 4 more than the work you perform, due to losses when glucose or fats are burned to allow the muscles to do the work. The Gibbs free energy change when glucose or fat reacts with oxygen and changes into water and carbon dioxide gives you the maximum amount of work that can ... 1 The final orbit has a smaller radius so we know the final PE must be less (more negative) and the final KE must be greater (moving faster in a smaller orbit) but the final total energy must be less (more negative), since$E_{total}=-\frac{1}{2}G\frac{Mm_{e}}{r}$. This means the satellite, or more correctly, the Earth-satellite system, has to lose energy. ... 7 Due to an accident of history there are two different units called the calorie and the Calorie - yes, the only difference is the capital C. The calorie is 4.2J but the Calorie is 4.2kJ, and the calories counted in diets are actually Calories even though they are invariably written on the food packaging with a small c. So you only used 3 Calories walking up ... 0 G is actually not independent of everything. It does depend on factors like density of space, rate of universe expansion etc. Its just scientists have not found this dependence. For time there have been many constants which were later were found to be dependent. For example at time of Kepler$T^2 =k R^3$Later, k was found to depend on Mass(by Newton) ... 0 A straightforward way to test your answer is to use a Monte Carlo method to estimate it. The Matlab code below evenly picks random points within two triangles that are both symmetrical about the x-axis (simplifies code because the net gravitational force must act along the axis, but it could be easily adapted to allow other configurations). It calculates the ... 0 The acceleration due to gravity on the surface of Earth can change in various ways as the earth shrinks. If you look up at the rate of change of g with respect to decrease in R the radius of Earth, it will vary as 1/Radius of the old value ; so the acceleration due to gravity increases with reduction in radius. If you also consider the change in spin ... 0 It is acceleration, change in direction in this case, which makes the difference as does the fact that graviton all attraction is a non-contact force. In space away from any large mass going at 1000m/hr in a straight line does not require a force to be acting on you. When you go around a corner it is a localised contact force that provides your centripetal ... 0 The value of acceleration due to gravity varies inverse squarely as the distance from the center of earth to the center of gravity of the object. Suppose we are concerned about the value of acceleration due to gravity on the earth's surface. Then the distance between the object and the center of earth will be the radius of the earth (assuming the earth to be ... 3 The strength of the gravitational field depends on how far you are from the center of the mass, assuming the mass density is radially dependent or constant. It also depends on the total mass inside your position: $$g=\frac{GM_E}{r^2},$$ where$r$is the distance from the center of the mass distribution. So if you stay where you are now, the gravitational ... 4 If you stayed at the same radius while the earth shrinks then nothing will change other than you'll start to fall. If you stand on the surface of the new smaller radius then you will feel the increased gravitation. 0 The gravitational forces acting on the two objects due to their interaction when 5 m apart have magnitudes of approx. 2.7e-2 N, giving initial horizontal accelerations of approximately 2.7e-8 and 2.7e-6 m/s^2. In the 1.49 s it takes for the two objects to reach Earth, those accelerations will change negligibly. The smaller mass will travel along a very ... 2 That's a very confusing paragraph. They use the term 'octave' specifically for a doubling of distance (r) --- that's not important. The argument is simply that the force per star falls off as$r^{-2}$, but the number of stars actually increases as$r^3$. So as you increase distance, the gravity increases by$r$. We could write this more precisely as for ... 2 There is one essential thing you have to keep in mind there is something important about physical formulas ...... think of the OHMS law R=V/I as you all know resistance of something is influenced just by its internal construction like the length,substance it is made from...etc since R is not influenced by I(current) and V(voltage) if you double current ... 2 First draw a circular path. In order for an object to follow that path at a constant speed, there must be a force acting on it towards the centre of the circle. At any instant, that force has no component in the direction of motion, and so, if that's the only force acting on the object, its speed will stay constant. Now draw a spiral path (i.e. one in which ... 0 If the gravity is caused by an infinite plane of mass, then yes: the equivalence is exact. In this case, Einstein original viewed gravity as a consequence of gravitational time dilation. Enter Earth: now the apples' paths cross--much like 2 nearby people walking on great circles around Earth. Thus, he realized the metric of space-time is curved. 0 I have tried to illustrate the error which has been made about the second term. Assume that a mass at position$A$($\vec r_1$) is under the influence of an attractive central force which originates from point$C$. The velocity of the mass is$\vec v_1$with radial and tangential components of velocity$\vec v_{1r}$and$\vec v_{1\theta}$. A little ... 0 The reason for the two behavior of the same objects accounts for the forces acting on it. When you drop a 5 kg body and another 25 tonne body vertically downwards, they will fall under the influence of gravity alone and both will fall with the same acceleration g (free fall). In such a case the weight of the body is zero. You know that fact, which you may ... 2 All other things being equal, if a heavier object will roll at a higher speed down hill than a lighter one With the qualification "All other things being equal" your statement is not correct. The falling acceleration is the same because doubling the mass of an object doubles the force causing the acceleration (the object's weight) which means that ... 0 Newton's constant,$G$, was introduced in his law of gravitation, $$F = G \frac{M m}{r^2}$$ In other words, the constant may be expressed as $$G = F \frac{r^2}{M m}$$ and the dimension of$G$may be deduced in the same manner as the dimension of velocity in$v=d/t$. Of course, if Newton's law of gravitation holds, all estimates of$G\$ from this inverted ...

0

Speed doesn’t depend on distance and time. We measure it by measuring distance and time. We should understand the difference between physics and mathematics. We use mathematics because we want to live better in nature. Speed is a abstract concept that we created it. We never can discover how the nature works or what are the rules nature based on? About G, we ...

1

if a heavier object will roll at a higher speed down hill Free fall and rolling are two different behaviors of objects. It is correct that for free fall all objects get the same acceleration ( minus friction and drag) but free fall is not the same as rolling. For going down a hill free fall can be compared to sliding, as was pointed out in the comments ...

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