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12

The answer is that inside a spherically symmetric shell of matter (your hollow earth or massive beach ball) there is no gravitational force anywhere - you will not "fall" in any direction, whether you are at the centre or not, regardless of the radius of the sphere. This is a classic result of both Newtonian Gravity, and Einstein's General Theory of ...


21

If the mass/charge is symmetrically distributed on your sphere, there is no force acting on you, anywhere within the sphere. This is because every force originating from some part of the sphere will be canceled by another part. Like you said, if you move towards on side, the gravitational pull of that side will become stronger, but then there will also be ...


0

Newton said that ever action has an equal and opposite reaction. Because of this, charges must either repel or attract, but cannot do an in between (This way the reactions are opposite. If it doesn't seem intuitive draw arrows.)


0

Answering your three questions: He knows the relationship between radius and terminal velocity, but the drops are too small to measure their radius with any accuracy (1 ┬Ám is tiny - looking at such a drop from a distance makes it no more than a speck of light, even with a "micro telescope"). Meauring the terminal velocity, he can deduce the size ...


2

Assuming that you mean $R_3$ is the initial distance from one object to the other, and given that gravitational force goes as the inverse square of the distance $F \propto \frac{1}{R^2}$, it follows that if $R$ increases by $3R$ (making the final distance $(3+1)R = 4R$), then the force changes by $\frac{1}{4^2} = \frac{1}{16}$ - it does indeed become 16x ...


0

Do you mean that the field is created by a stationary charge $C$, and $r$ is the distance from your particle to it? Even then, if you want acceleration you have to multiply your expression by $q/m$, where $q$, $m$ are the mass and the charge of your particle, respectively. More generally $a=\frac{q}{m}\,E$, where $E$ is the electric field strength. See ...


0

The parachute can not be opened instantaneously. The parachute is fully opened at the "vinicity" of the point $C$. The acceleration due to gravity is constant but there is another force one needs to take into account : the air resistance.


2

In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific ...


0

Assume a person is falling towards the earth. We know that there is a force and thus an acceleration acting on the person. The opposing force is the gravitational force exerted by the person onto the earth equal in magnitude (Newton's Law of Gravity). This force produces an acceleration (Newton's Second Law) but because the mass of the earth is massive as ...


1

Here is one way to solve it. You can multiply both terms by $\dot r$. $$ m\frac{d^2 r}{dt^2}\frac{dr}{dt} + \frac{GMm}{r^2}\frac{dr}{dt} = 0 \quad\Longrightarrow\quad \frac{d}{dt}\left[\frac{m}{2}\left(\frac{dr}{dt}\right)^2\right] - \frac{d}{dt}\frac{GMm}{r} = 0 $$ Basically $\dot r$ is the integrating factor. We now recognize the potential and kinetic ...


1

You seem to be asking for a higher order approximation to the acceleration due to gravity based on its variation with height. So if the Earth's mass is $M$, and we're at a radius $R$ from its centre, then for a small radial displacement $h$ from this, the acceleration due to gravity is: $$g = \frac{GM}{(R+h)^2} \approx \frac{GM}{R^2}\left(1 - \frac{2h}{R} + ...


2

If the person in concern is standing perfectly, such that his weight distribution is equal over both legs, and if both the weights are calibrated perfectly, then yes, both the weights will show equal readings. (The readings each being half the weight of the person.) In a realistic case (without 100% perfection), however, the weight would be unequally ...


1

$N_1$ the reading of force on the first weighing machine $N_2$ the reading of force on the second weighing machine $X_1$ the horizontal displacement of first leg from COM $X_2$ the horizontal displacement of second leg from COM we know that the man is in equilibrium. so the weights shown on both meters will have their sum equal to ...


1

Remember Newton's first law $\sum F=0$. Vertically you have this: $$\sum F_y=N_{scale1}+N_{scale2} - W=0$$ The scales will feel and display the $N_{scale}$ force. Solving for either scale force you will see that they share the total weight $W$. Together they must lift the whole weight $W$, not individually.


2

I just want to formulate John Rennie's answer in a slightly different and more general way. For any object of angular momentum $I$, the angular acceleration $\dot \omega$ is given by $$\dot\omega = \frac{\Gamma}{I}$$ where $\Gamma$ is the torque. Now in the case of an object with masses $m_i$ distributed along the length at distance $\ell_i$ from the ...


1

An easy way to explain it is a light source emitting light in all directions. The light is the area of a sphere, and volume has nothing to do with it.


2

This, in fact, is not too different from the flat surface case, the difference lies only in the distribution of the force components along the '$x$ and $y$' components of velocity. In the flat surface projection case, you have acceleration only along the vertical direction. The given case can be converted to a non-inclined plane case by mentally rotating ...


1

Flux is proportional to the area of the sphere not the volume of the sphere. It is evident from definition of the flux $\Phi_\mathbf{B}$ of some quantity $\mathbf{B}$ , which is defined in the following way, $$\Phi_\mathbf{B}= \iint\mathbf{B} \cdot \mathrm d \mathbf{A} $$ Therefore the flux is proportional to the area of the sphere and hence the $1/r^2$ ...


1

All the stars would be attracting each other and hence the would stick to each to attain equilibrium. Why doesn't this happen? You are forgetting angular momentum. Consider a binary star pair. Ignoring the expansion of spacetime, and in the absence of some mechanism that removes angular momentum from the system, those stars will orbit one another ...


0

When you analyze motion problems using Newton's Laws you draw a free body diagram for each body you want to analyze. Once you have done this you choose a coordinate system in which you will write Newton's 2nd law equations. I believe you are having a fundamental uncertainty about how to choose that coordinate system. For the simple pendulum, you see the ...


0

Let me summarize the discussion in the comment section. First of all you can safely use Newtons equation of motion $$ F = m g $$ with a constant gravitational acceleration $g$. This is simply because the height of a 10 stories building, lets say $30 m$, is still very very small compared to the radius of the earth which is about $6000 km$. Therefore, as ...


4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...


-2

F=ma is saying that if a mass 'm kg' is accelerated at 'a metres per second per second' then there has to be a constant force of 'F' pushing it. a better equation to represent the effects of gravity would be: F=(G*m1*m2)/d^2 where: G= universal gravitational constant (6.6738410^-11 m^3 kg^(-1) s^(-2)), g= acceleration due to gravity (9.81 m s^(-1) ...


2

What if this planet existed and you fell in? Let's take a look at that and see. I'll spare listing all the reasons why a planet like this can't exist and wouldn't last long if it did; those are no fun. So instead, let's say this planet does, for some reason, exist in some galaxy far far away and it's made of something strong enough that it'll be around for ...


1

All the stars would be attracting each other and hence they would stick to each to attain equilibrium. Why doesn't this happen? This is an old question. Even Newton himself had thought about this question. His idea was that in long distances or separations (say, inter-galactic distances) the force of gravity might appear to be repulsive. That's why not ...


47

I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like: The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity. To write down the equation of motion ...


7

I assume that you're imagining two sticks whose bases stay stuck still to the ground as they tip over. In this case, what we should do to calculate the tipping rate (I wouldn't exactly say "falling"-- I think it's misleading) is consider the torque applied to each stick as it tips. What the answer really comes down to is the struggle between torque (the ...


-3

Yes you are right. There is a resist in gravity because of an applied weight. Therefore the acceleration of the beam falls slightly slower depending on the mass of the object. Unless nothing are holding them together, both beams would fall at the same rate except the applied weight on the top would fall slower. For an example, If we are in an elevator and ...


-1

Take a baseball and put it at about ground level and drop it into the hole. The gravity will pull on the ball and will accelerate and plunge through the planet, the void, and on it's way up again (or down from your perspective), gravity will slow it's ascent to the point that it'll just about make it to ground level on the other. The ball will spring back ...


-1

If there would be initial rotation(as we see most of all objects today are rotating around another) gravitational force is accounting for the centripetal force. I'm new here so, I don't know how to type the equations, but I hope you get my point. Further many objects are there which have many other forces, like Coulombic force(when charged bodies are ...


6

Emilio Pisanty's answer is a great one to this particular answer, i.e. there is no in principle bar from the laws of physics to a structure like yours and it would have the zero gravity inside property that Emilio describes. But, from a materials perspective, it would be pretty much impossible for a planet like this to form unless very small. The stablest ...


1

I have highlighted some key word lacking in your revision. Also, work has a very specific definition. The difference in gravitational potential difference between $\vec{r}_1$ and $\vec{r}_2$ is the negative of the work done on a unit mass by the external gravitational field as the unit mass moves from $\vec{r}_1$ to $\vec{r}_2$. As an example, consider a ...


1

Words are imprecise. Your wording is much less imprecise than the one from the textbook. As the work is positive or negative (depending on the direction taken), there is no need to define a "positive direction", but the potential difference depends on the order of $\vec r_1$ and $\vec r_2$ (that is: if the potential difference moving from 1 to 2 is positive, ...


4

According to the shell theorem, the gravitational force inside such a hollow spherical planet would be 0 On the outside, gravitational force would be as if the planet was a regular planet. At some point during your fall, the gravitational force pulling you to the center would decrease to 0. If there's atmosphere inside the planet, it would slow you down ...


21

Yes, this is possible. It is perfectly fine for a mass configuration to produce, for points outside a sphere of radius $R$ centred at $\mathbf r_0$, a gravitational field identical to that of a point mass at $\mathbf r_0$, and still be completely empty inside a smaller sphere of radius $a$ around $\mathbf r_0$. The spherical-shell model you describe is ...


3

In your context, the second interpretation is correct. The fact is that falling objects accelerate both on Earth and on the Moon. The sentence is saying that the amount of this acceleration, regardless the source, is six times greater on Earth than on the Moon. In other words, things accelerate towards the surface of Earth six times faster than they ...


0

The leading theory for formation of the Moon is that a large object collided with the Earth and threw off a cloud of debris. This cloud then clumped together under it's own gravity to form the Moon. Simulations of the cloud show it formed outside the Roche limit - at around 1.3 times the Roche limit in fact. So the Moon has never been closer to the Earth ...


0

When the body is falling down without the affect of air resistance,it will experience only one force,I.e.the gravitational force which pulls out downwards towards the earth..and the object will fall to the surface of the earth will an acceleration 'g'. Coming to the first case...when the object is falling down in the presence of a drag force (friction due ...


1

In physics, escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero.[nb 1] It is the speed needed to "break free" from the gravitational attraction of a massive body, without further propulsion, i.e., without spending more fuel. For a spherically symmetric massive body such as a ...


0

Potential has a negative sign as work has to be done against the external force to bring the mass from infinity to the point (definition of potential )


4

Because of a convention wherein zero gravitational potential is said to be at infinity. See Wikipedia: $V(x) = \frac{W}{m} = \frac{1}{m} \int\limits_{\infty}^{x} F \ dx = \frac{1}{m} \int\limits_{\infty}^{x} \frac{G m M}{x^2} dx = -\frac{G M}{x}$ "By convention, it is always negative where it is defined, and as x tends to infinity, it approaches zero." ...


1

Hmm. I think there should be a minus sign and it doesn't matter where you set the zero of potential. Gravitational field strength is the negative gradient of the potential. For a spherically symmetric field $$ g(r) = -\frac{dV}{dr}.$$ If $$V = -\frac{GM}{r} + V_0,$$ where $V_0$ is an arbitrary constant, then $$g = -\frac{GM}{r^2}.$$ i.e. your book is ...


3

The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ...


0

The period for normal pendulum is proportional to (L/g)^1/2 so you replace g with gcos(30) the new restoring force is cos(30)=(root(3)/2) so period will increase by factor root(2/root(3))


1

Well, if the string was pulled such that it was along a radial arm from the centre of the "earth", in other words, in a vertical, then it wouldn't sag. Any deviation such that it is no longer exactly vertical, then the above answers come into play. I know this isn't what you are really asking, but let's admit, it does apply to the question as stated.


2

As per @lemon mentioned : You can treat the sphere as a single point (this follows from point 1 of the shell theorem). So you just need to integrate along the length of the ring. So I just did like that only. Taking sphere as point mass then point mass will experiences one gravitational force of attraction that is: $$df=cos\theta$$ As vertical ...


-3

Motion The fundamental relationship between Coloumb's law and Newton's Law of Gravitation is Motion. To explain let us first take Einstein's E=MC2. Now we know that anything times one is itself. Therefore when mass is 1, E=the speed of light squared. Therefore we can see that each piece of mass is the storage of the speed of light squared. Now in geometry ...


3

Cosmic Velocity has nothing to do with infinity. A cosmic velocity is the minimum speed directed in the necessary direction to escape the gravitational attraction of a cosmic body such as a planet, a star, or a galaxy. Here is a paper which a student wrote about the four cosmic velocities. I don't know if his exact classifications are in common usage, but ...


2

Gravitational force is really weak compared to the other fundamental forces, so it's very difficult to measure the gravitational constant. This is how Cavendish did it without knowing the Earth's mass: He put two lead balls on either end of a long bar. He hung the bar at its center from a long twisted wire with known torque. Then, he placed two really ...


0

Brionius dealt with the value of G. Cavendish's experiments also confirmed the product of masses term. The inverse-square portion of Newton's Theory of Universal Gravitation was immediately accepted, since it straightforwardly produces Keppler's Laws. Further thought shows that, in order to produce stable orbits, the exponent must be exactly 2 - no more, no ...



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