# Tag Info

1

Let's use a spring scale in water. Putting both the feather and the iron ball of same masses we notice that the spring shows the same reading(say $a_0$) for both feather and iron ball. Hence the gravitational force on both the feather and iron are same. Now let's repeat this experiment in air(or oil or some other material). We notice that the spring scale ...

0

The mass $M$ in the equation for orbital velocity would be equal to the mass of the Earth (assuming that the mass of the satellite can be neglected) and thus constant. The gravitational constant, $G$, like the name suggests is constant as well. But the radius of the (circular) orbit, $r$, can vary. So by using different altitudes at which the satellite would ...

2

As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.

1

Gravity on Earth uses two concepts, one is the radius (R) of earth and the other is the distance (h) from the surface of the Earth. Really, approximating the Earth to be spherical and uniform, it is just one distance, the distance from the center of the Earth that matters. $F = G \frac {M_{Earth} m}{r^2}$ where $r$ is the distance from the center of ...

0

If you have ever seen a pizza being made by hand, you will know that when the baker throws the disk of dough in the air, he makes it spin. As he does so, the pizza "disk" gets bigger because the dough on the outside experiences a larger centrifugal force (in the rotating frame of reference of the pizza. Don't start on "there is no such thing", you asked for ...

1

There is a wikipedia article which describes the effect http://en.wikipedia.org/wiki/Equatorial_bulge Basically the bulge is caused by the rotation of the Earth. The centripetal force is given by $F=m\omega^2 r$. Therefore the poles feel a lesser force than the equator which wants to spin out into a disc. This is balanced by gravity which wants the Earth to ...

1

The answer is "maybe". Anomalous gravitational effects in black holes just exist beyond a region called "event horizon". After the collapse, only bodies moving very closely the "event horizont" can feel these bizzare gravitational effects. That's a good example about it: if the sun collapses to a black hole right now, gravitationaly, nothing would change for ...

1

Here are several thought experiments (and what happens in each). I'll ignore relativistic effects like time distortion - not for your sake, but mine :) The earth collapses to a black hole beneath our feet. We fall with it, and end up inside a black hole, presumably dead. The earth collapses to a black hole beneath our feet, but we stay in the same place. ...

2

The formula $F=G \frac{m_1 \cdot m_2}{r^2}$ is valid only for point masses. However, it can be applied to non-point masses if its spherically symmetric. Enter Shell Theorem: 1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. So, when a spherically symmetric ...

4

During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly. Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) ...

1

The equation for gravitational force F=Gm1m2/r^2 gives the force of attraction b/n any 2 bodies with point mass m1 and m2 and separated by a distance 'r'..it means both the objects are attracted towards each other by a force F=Gm1m2/r^2..It is also coherent with newtons 3rd law i.e action and reaction forces are equal.. The expression F=ma or a=F/m ...

0

Yes, this is all correct so far. What you need to remember here is that Force is a vector quantity. That is, it has a direction associated with it. A force pushing you into the ground is the not same as one pushing you up into the sky, like the seat of a flying airplane. So here you need to lable your force $\vec{F}_1$, say, and this would be the force ...

1

If $a$ is the acceleration of object 1 (should write as $a_1$), then $m$ should be $m_1$. Vice versa.

0

First of all, I would like to say that the answer Floris gave is the correct way to do the problem you've set forward, but I thought it worthwhile to note that the result ${5 \over 2}(R-r)$ is the answer if, rather than rolling a marble down the track, you are sliding a cube. If this problem comes from a textbook, there were some unstated assumptions in ...

2

Your expression for the velocity looks right; but we have to get a few other things taken care of. First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression. On the other hand, you need to take account of the energy of the sphere ...

0

In classical mechanics the gravitational force does not have a spacial limit, i.e. any mass no matter how far it is from the earth will experience its gravitational pull (and vice versa). However, this does not mean that any object thrown into space will eventually return to earth. This is due to the fact, that the gravitational force on an object becomes ...

0

Yes, this is exactly the definition of the escape velocity. I cannot remember the number for it, but I'll trust you that it is 11. Either way, you derive this by doing an integral, where you take the final position at infinity. That is, you assume that the particle that you shoot up into the air/ space goes off forever and never comes back. It follows ...

1

I assume this question is about classical mechanics. Therefore gravitational force doesn't have a limit. According to conservation law of energy (P.S. Gravitational energy $E_p = -\frac{GMm}{r}$): $\frac{1}{2}mv^2-\frac{GMm}{r}=0,\ when\ r\rightarrow\infty$ Therefore, the escape velocity is $v=\sqrt{\frac{2GM}{r}}$.

4

Earth has few and relatively tiny sattellites other than the moon specifically because of this large moon. Note the mass ratio of our moon to our planet. It is the highest in the solar system by a large amount. This one large sattellite will over time sweep up and aggregate other smaller sattellites. Put another way, we probably did have other smaller ...

-1

The moon is by far the largest of earths natural satellites, but is by no means the only one. There are many much smaller bodies which either orbit earth or have an orbit around the sun which is extremely similar to that of earth. The moon is the most significant and largest as it is the result of large amounts of matter clumped together that was thrown off ...

-1

When the the earth was formed. I guess only the moon was close enough in proxmity to be attracted by the gravity of the earth or rather earth cant hold anymore natural sattelites but the moon.

1

Your answer (1) is the correct one. It is actually quite simple if you think in terms of conservation of energy. What you have described is a simplified version of a two body problem. Note that strictly speaking, both the doughnut $(D)$ and the ball $(B)$ will move towards each other. But without outside influence, their combined center of mass should be ...

3

The gravitational field $\mathbf g$ equals, by definition, negative of the gradient of a correspondonding potential $\Phi$; \begin{align} \mathbf g = -\nabla\Phi. \end{align} Therefore, it suffices to produce a gravitational potential $\Phi$ whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do. Let a ...

0

Newton's law of universal gravitation holds for point like masses. For spatial masses you would have to integrate of infinitely small parts of that mass. However it does turns out that the result of this for a sphere of constant density at a given radius does yield that simple formula, see Gauss's law. To get a better understanding of this, without having ...

0

when your ball is far from a donut, there's a potential energy of the gravitational field. when the ball reaches the donut, that potential energy must be converted into something else, it can't disappear. do it must be that it converted into the kinetic energy of movement, i.e. the ball must be moving at this point, hence, it'll pass the center at some ...

1

1) is correct. The wrong reasoning about 2) is that what you have in mind is probably Newtons Law for point masses. When the sphere is close to the doghnut the gravitational force will be more complicated, but still point towards the centre of the doughnut due to the symmetries in the situation. It will however stay finite, because all points of the sphere ...

1

There follows my try to decompose the solution into a minimal amount of calculation and apart from that only geometrical considerations. The centripetal acceleration is $a_c=\frac{v^2}R$. It is directed towards the center. We define the $z$ coordinate as starting at the top and pointing vertically downwards (see the following Figure). The conservation ...

1

The harsh answer is, "by solving Einstein's field equations". That is an extremely more difficult problem than using Newton's law -- the equations take up at least a page when you write them out. However for the case of light near a black hole, you can treat the black hole as unaffected by the light, and instead solve the null geodesic equation. This is ...

1

A mass moving in a circle has centripetal acceleration $v^2/r$ directed toward the center of the circle. You can get $v$ from potential energy. The mass here has two forces on it. Gravity is constant and down. The reaction force of the surface (assuming no friction) is normal to the surface. When the sum of these two forces becomes less than centripetal ...

0

I think your definition of sphere of influence is not correct. You could also be confusing the sphere of influence with the Hill sphere. The sphere of influence has mainly an application in the patched conic approximation. And the word sphere is even another approximation. A related question asks about the derivation of the radius of this sphere. I also ...

0

I hope this doesn't confuse you, but in one sense, yes, heavier bodies do fall faster than light ones, even in a vacuum. Previous answers are correct in pointing out that if you double the mass of the falling object, the attraction between it and the earth doubles, but since it is twice as massive its acceleration is unchanged. This, however, is true in the ...

-1

Although Pranav Hosangadi has explained, I will try to explain where you might be going wrong. I think this will be helpful for you. And I think it will not be waste of time in typing the answer for you. I was going to ask: if mass is an objects tendency to resist acceleration then why do two objects of different masses fall to the Earth at the same ...

0

Seeing your comment, it seems you are concerned about group of charges with certain mass. Then you need to apply Gauss law for the cases where it becomes difficult to apply coulombs law or principle of superposition. In case of gravitational force, find the center of masses of either configuration and you can proceed to find force using Newtons law of ...

0

The magnitude of gravitational force between them would be: $$G\frac{mm'}{d^2}$$ The magnitude of electrostatic force, if the magnitude of charge on them is $q$ and $q'$ respectively, would be: $$\frac{1}{4\pi \epsilon _o}\frac{qq'}{d^2}$$ The net force would just be the vector sum of the two.

0

Since $k_e$ is of the order $10^{9}$ and $G$ is of the order $10^{-11}$ thus gravitational force can be neglected, even if you add then can be added safely thus $F_T = F_g + F_e = G\times\frac{m_1\times m_2}{d^2} + k_e\times\frac{q_1\times q_2}{d^{2}}$

-1

Mass is an object's tendency to resist acceleration. This applies when both masses you're testing are subjected to identical forces. From Newton's Law of Gravity, $$F = G \frac{M \cdot m}{r^2}$$ It is fairly obvious that the force the Earth exerts on a heavy body is more that what it exerts on a light body, so you can not compare the accelerations ...

2

A Lagrange point is a position relative to a system of two-body gravitational objects at which a third negligible small object, if its velocity is correct too, would not move relative to the two other objects. The two gravitational objects will orbit around the center of mass of both objects together (also called the barycenter). I am not sure, but I also ...

1

For your first question: Yes gravity is mediated in all directions equally. As for your second question: I assume you are asking about elliptic orbits. If this is the case then, from the law of conservation of angular momentum, when the planet is closer to the sun it has greater velocity, and when its further out it travels slower because $\vec{L}$ = ...

0

Consider a lift with its rope snapped. The lift would be falling freely. An observer is inside the lift (tough luck for him!) releasing the ball just at the moment of the free fall. Since the ball and lift would be falling freely the ball would appear to float. Thus, to the observer in the lift, it would seem as if no force is acting upon the ball, using ...

1

UPDATE I finally got it right. My thanks goes to @NeuroFuzzy, who pointed me in the right direction. According to wiki's Legendre polynomial solution for the elliptic integral, "an exact solution to the period of a pendulum is:" $$T=2\pi\sqrt\frac{\ell}{g}\sum\limits_{n=0}^\infty ... 1 Even though the people who replied are correct, fun fact: the speed that gravity is propagating is very slightly affected by mediums. Here are 2 pages from Kip Thorne's lectures at Les Houches. Gravitational waves are affected by a medium through a dispersion relation, but the effect is ridiculously small and is neglected. 1 There are multiple forces involved in your thought experiment. Namely, the gravitational force and the 'force' of air resistance. This does not mean that the presence of the air affects the gravitational force. As another thought experiment, consider two planets orbiting in space. Somewhere between these two planets there is a point where the ... 2 To analyze whats going on you need to sum all forces that affect the objects in question. For each object, the feather and the ball, the forces of gravity are identical. The other forces are not. 3 Your thought experiment of dropping an iron ball and a feather need not be in water; in fact, it is more commonly considered in air, but the pertinent facts are the same. All objects, regardless of their mass or composition, are accelerated identically by gravity. But within a particular medium, the acceleration of particular objects might be impeded by ... 2 As you said, if feather has a stronger decelaration, is due to air friction (not because iron is more dense than feather or something). The same should hold in absence of gravity: if you give a boost to these balls, then the feather one will stop first. From this last consideration, I'd say that this medium does not affect gravitational force, but just it ... 1 This is really a comment to Mathbreaker's answer, but it's hard to do formulae in comments. If you simply solve:$$ T_m = 2\pi\sqrt{\frac{\ell}{g}}(1+\frac{1}{4}\sin^2(\tfrac{\alpha}{2})) $$for g you get:$$ g = \frac{4 \pi^2 l (4 + \sin^2(\tfrac{\alpha}{2}))^2}{16 \tau^2} \tag{1} $$We use the identity:$$ \sin^2(\tfrac{\alpha}{2}) = \tfrac{1}{2} - ...

0

$$g= \frac{L *4\pi^2 * ((5- \cos^2 (\alpha/2))^2}{16T^2}$$ This is what i got when i simplified for $g$. I assume you made some mistake in simplification, or it just might be me, Iapologize if it is. You could try plugging your values into this and hope for an answer... when i simplify the latter part as well i get $$\frac{L * 4\pi^2 ... 3 It seems like they were able to rigorously prove the existence of N-body choreographies by using interval Krawczyk method to show that a minimum exist to the variational problem solved in the subspace of the full phase space satisfying some symmetry conditions. Following the links given I found this paper where they explain the method. It's not exactly a ... 0 What happens to the force of gravity as the separation between two masses approaches 0, is that it increases. Although it can attain a large value, in practice, it never goes to infinity. This is due to the fact that the masses, even if very small, have a radius. Therefore, the closest they can be is r1 + r2, were r1 & r2 are the radii of the masses m1 ... 0 The "intuitive" expectation is that both methods should give the same answer - and they do, if you do it right. Were method 2 goes wrong is in calculating the force at distance$$d = \frac{r}{\cos{ \alpha}} although this gives you the force between M and m, this is the wrong force. The force that is required, is the force along a line between m and the ...

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