# Tag Info

-3

The gravitional constant is so small because before there exists a force, there are some substances which compute together to form forces, so the sum total of the masses of there atoms, which is very small, gives the value of the force constant. You can prove this theoretically.

0

You want to integrate with respect to $t$. You have $\ddot x = g(1-\frac xR)$, which is a harmonic oscillator equation. With the initial conditions, it won't oscillate, but you can use the usual solution technique. Your choice of $x$ as depth instead of radius complicates things-I suggest using radius $r$ measured from the center so $\ddot r=-gr/R$ Find ...

0

In one dimensional motion, sign really does mean direction for vectors. You can say positive is upwards and negative is downwards or vice versa. Now, come to your case. You seem to have mathematically understood. Then start with making your intuition. As stated by other answer, suppose you are very,very far away where gravitational force of the earth ...

0

Here is a simple model as explanation: Imagine yourself far away from any gravitational field - your potential energy is zero. As soon as you are entering into a gravitational field, you are accelerating and winning kinetic energy. The origin of this energy is that you "borrowed" some energy which is the potential energy you are losing. When you want to ...

2

Unfortunately the answer to "is the structure gravitationally stable?" is "most definitely not." Anything planet-sized pretty much has to be close to a sphere, unless it's spinning very rapidly, because the gravitational forces increase with the body's size, whereas the electromagnetic forces holding atoms together don't, so the material's strength will ...

1

Going under water would have a similar effect, thats how astronauts train for performing operations in 0 gravity like space walks ect i believe.

0

Remember the definition of work: $\vec F.d\vec x=Fx\cos\theta$. In your case $\theta=180^o$, thus $\cos\theta=-1$. That is the minus sign you were missing.

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In your formula you don't calculate the potential energy, but the difference between the potential energy of the body at the lower level $P_E(h_0)$, and that at the higher level $P_E(h_1)$. Indeed, since $P_E(h_0) < P_E(h_1)$ you get $P_E(h_0) - P_E(h_1) < 0$ as expected. Phenomenologically, starting from $h_0$, you have to invest some work to raise ...

0

Gravitational acceleration at the surface of earth varies with latitude (North-South position). This is due to 1) the outward centrifugal force produced by Earth's rotation, and 2) the equatorial bulge (itself caused by Earth's rotation). Both effects cause the gravitational acceleration to decrease away from the poles. The net effect is a gravitational ...

0

The value of acceleration due varies with altitude as well. It actually decreases with increase in altitude. The following formula approximates the Earth's gravity variation with altitude: $g_h = g_0\left(\frac{r_\mathrm{e}}{r_\mathrm{e}+h}\right)^2$ where $g_h$ is the gravitational acceleration at height $h$ above sea level. $r_e$ is the Earth's mean ...

0

You can calculate the value of acceleration due to gravity in your town by finding the latitude of your town. Once you find the latitude use this equation $g'=g-R\omega^2\cos^2\lambda$. Here $R$ is the radius of the earth, $\omega$ the angular velocity of the earth and $\lambda$ the latitude. This equation comes from the fact that the earth is rotating and ...

2

From my knowledge, gravity is infinite and extends throughout all of space. It diminishes as distance increases but is still present everywhere. So given enough time, no matter where something is in the universe, it would accelerate due to the gravitational force of something in the universe. You must mean "the effect of gravity reaches to infinity" . ...

1

The air density of your planet is on the order of $10^{-18}$ of air at the surface of the earth. Gravity Force would be 10 times larger, even accounting for a slightly larger radius. The largest possible parachute that could fit around such a planet would be in the order of $10^{12}$ times bigger than a normal parachute. But it would need to have a mass in ...

1

From the equation given for terminal velocity v=$$\sqrt{\frac{2mg}{\rho AC}}$$ the density of the planet in question will be very low , so any object dropped will take a very long time to acquire a terminal velocity. There are so little particles present in the atmosphere that the balancing of the weight by the drag and buoyancy will take a very long ...

0

I don't deal with drag forces often, but I think the equation for drag is $$F_D=Cv^2,$$ where $F_D$ is in the same direction as $v$, and $C$ contains all the various things - density of air, cross-section, drag coefficient, etc. Importantly, $C$ depends on the orientation of the object. What I am going to do is assume the bullet falls without rotating - so ...

12

Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground. Just consider the vertical force caused by the air friction: $F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$ Where $\theta$ is the angle above the horizon for the bullet's velocity, ...

1

Let's make a few simplifying assumptions. We'll use only one ball, and assume that when the ball returns to your hand you immediately throw it back up - we'll assume the time taken for the throw is short enough to be ignored. We'll also assume your arm isn't elastically storing energy, so for each throw you have to put in fresh energy from your muscles. ...

1

In order to understand how gravity affects objects at different points within or on a sphere, see the following link: http://en.wikipedia.org/wiki/Shell_theorem This basically states that inside a sphere, the gravitational force of shells further away from the centre than you cancel each other out. Considering your initial question about pressure at the ...

0

It may help: suppose we are close to the earth and at height $h$. So $$\Delta V=Gm_1m_E(\frac{1}{R}-\frac{1}{R+h})$$ where $R$ is the radius of the earth and $h \ll R$. Now we approximate this relation and it's turn out that $$\Delta V=Gm_1m_E(\frac1R-\frac1R+\frac{h}{R^2})$$ By calling $g=\dfrac{Gm_E}{R^2}$, we find $\Delta V=m_1 gh$. Even if we don't ...

4

You are correct that as you get very close to the center of the earth, the value of $g$ can become arbitrarily low. If you could somehow create a space there, you could potentially float in it because you would not be pulled in any particular direction with respect to the earth. But while gravity is not strong there, it is strong in other places (like your ...

1

I would guess that as long as they are both constantly being juggled that it would be the same, because throwing a ball in the air at twice the speed will result in twice the time to fall, but also twice the work to achieve the speed. I'm just using deductive reasoning and some of the basics i have learned. I hope this helps!

0

You have misconcieved the expression for work in your question as you've assumed $W$ as the total work on the object. It is actually $W_g=\Delta K + \Delta {P.E}$ ,where $W_g$ is the net gravitational work on the object. You see, $W=0$ while $W_g=-mg\Delta h$ in your case. And thus, $\Delta P.E=-W_g=mg\Delta h$

0

If we estimate the size of the Earth to be a perfect sphere (which it isn't, but as a first approximation it will do), then you may apply the shell theorem. It states that the gravitational field of a spherically symmetric body appears as if it is concentrated in the center of mass of the body. Knowing how much Earth weighs ($5.97219\times10^{24}\,kg$), you ...

0

Let us assume, you lifted the book infinitesimally slow and the velocity is zero so is the acceleration making change in kinetic energy zero but the work is never zero. It is mgh(m-mass of book, g- acc. of gravity, h-height to which the book is lifted). The work done by you carrying the book around, that will be zero since no change in height or speed of ...

1

There are several possibilities for your confusion - and you bring up several related concepts (Work, potential energy, kinetic energy). When this happens I think it's helpful to isolate one thing which you know must, absolutely be true. The definition of work (for a constant force) is $$W=\vec{F}\cdot \Delta \vec{x}=F\Delta x \cos\theta.$$ First ask "what ...

0

Doing work on an object will increase its kinetic energy, but it could also increase its potential energy and therefore not increase its $KE$. To raise a $4kg$ object by $25 meters$ you have to counteract the force of gravity. If your doing this with a force equal to $F_g$ then when you've lifted the object there will be no increase in $KE$. To raise a ...

2

The case on the left is correct. Gravitational gradient goes in the direction of the source of gravitational force. Think about the extreme case: if the object was leaning almost all the way over at 89.9$^o$, the center of gravity would necessarily be almost directly below the center of mass.

2

If the density is constant, it is guarantedd that M and G will fall into the same vertical line. However, if the density is not constant, it is not difficult to find a counterexample that will have G and M on diffrerent vertical lines.

2

In our Moon, the center of gravity is not the same as the the center of mass. The result is that the Earth always sees the same side of the moon. This is becasue gravity pulls at the center of gravity, but the orbit is determined by the center of mass. The center of mass determines the kinematics - how an object will rotate, spin, revolve, and orbit. If ...

2

The water inside is a fluid, so it isn't rigidly attached to the walls of the bottle. This means that the bulk of the water will still accelerate at $g$, save for the part of the water close to the bottle walls, which will be dragged along with the bottle. The water isn't really rising up, it's just falling slower than the bottle. In the frame of the ...

23

As a quick rehash of layman's terms definitions you've probably heard: The Center of Mass (CM) represents a single point where you could treat the object as a point particle, with the combined mass of the object. It's found by the average location of the mass of an object. The Center of Gravity (CG) is a point that represents the average pull of gravity on ...

31

Both values are computed as a position weighted average. For the center of mass we average the mass in this way, while for the center of gravity we average the effect of gravity on the body (i.e. the weight). \begin{align*} x_{com} &= \frac{\int x \, \rho(x) \,\mathrm{d}x}{\int \rho(x) \, \mathrm{d}x} \\ \\ x_{cog} &= \frac{\int x \, \rho(x)\, ... 8 The centre of mass is the average point of the "mass" of the body, whereas centre of gravity is the average point of the "weight" that is mass times the local gravitational acceleration. For small objects both of them are almost same, but for large objects as the value of gravitational acceleration can change along the body (as the gravitational ... 1 You may be surprised to learn that you have done the right thing and have got the right answer. It's just that you've used non-standard units. Because you've put the distance in as millimetres and the time as milliseconds the value you calculate for g/2 is in units of millimetres per millisecond squared. You need to multiply it by a thousand to convert it ... 1 Let's start out by understanding what's going on when we take logaithms. We have the relationship y=kx^n $$There's no obvious method for working out either k or n from the graph of y against x; instead, we plot \log(y) against \log(x). Why do we do that? Well, if we take logarithms of both sides of the above equation, we get:$$ ...

-1

I believe the Earths Inertia would prevent any of the other 3 objects from moving it. Their Gravity can not overcome the Earths inertia. Therefore The Earth stay's where it is and the 3 land at exactly the same moment.

1

No, to translate escape velocity $v_{esc}$ at the surface of a planet into gravitational acceleration $g$ at the same location, you also need the radius $R$ of the planet. The equation that applies is: $2 g R = v_{esc}^2$.

2

It is true that the shell outside the current radius does not contribute, so you are only left with the force from the mass inside: $$F=G\frac{mM_{inside}}{r^2}$$ but the M inside becomes smaller as the radius becomes smaller.Assuming the dendity, $\rho$, is a constant, then $$M_{inside}=\rho\frac{4}{3}\pi r^3$$ which leaves you with: ...

1

As a starting point for my answer I take this comment by bobie under his question: ... my goal is to ascertain if Galileo could in his age prove that g is the same for2 bodies as different as a feather and a hammer, without cheating ... The problem in our experiment is air. If it wasn't for the air, the experiment would show unequivocally whether a ...

2

It seems to me you may be misunderstanding the problem as stated. You are assuming you are being asked about two different objects (planets?) in different orbits; but I think from reading the question that you are being asked about the same object at different points in its elliptical orbit. For an object in an elliptical orbit, conservation of angular ...

3

Your equation relates the period of the orbit to the length of the semi-major axis, not to the absolute distance at any point. You can use the Vis-viva equation if you have more information. But you don't have the semi-major axis length or other details about the orbit. As you suggest, conservation of energy is the simpler way forward.

1

Two issues: 1) the equation was derived assuming that the initial speeds were zero, so both masses started at rest. To get a more general expresion yuo need to integrate again (I'll check later if this is easily doable) 2) you can apply this equation in 3d, the equatiosn assumes that the masses start at rest and follow a staight path until they collide. ...

10

One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be ...

0

Make the hammer the same weight as the feather (I know it would be small, but it would be a hammer) Get a bigger feather. Scrunch up the feather into a ball (less resistance). Stick the feather under the hammer.

1

It is not practically possible that the feather does not touch the box. I mean, even if you remove all the air inside the box, gravity would pull the feather to the base of the box BEFORE starting the experiment. If you clamp the feather to the center of the box, your experiment would get biased. I suggest working on the shape of the competing weight (the ...

3

For a heavy enough hammer, and not too high distances, the friction from the air will not be too large, because it is a function of speed, so the smaller the speed and the higre the mass the smaller the effect of the air friction. But even if there is some friction this should not be a problem (as I will explain at the end). If in your experiment the cage ...

2

1) What is the initial position of the bannana? 2) invent an initial position for the shooter. 3) at time 0, what is the direction of the bullet's velocity? 4) Now, set up a system of equations for the bullet and the bannana's motion. 5) do they hit each other?

10

Yes, Newton's formula is just fine. No, the formula in your book doesn't describe reality. At first this sounded like an exercise, where the next sentence is probably something like "calculate the effect this has..." These sorts of hypothetical questions are meant to show you how you could distinguish between competing physical theories. Some more digging ...

2

In principle not, because it is wrong when describing the behavior of the orbit of Mercury, but should be borne in mind that there is no absolute truth when describe the universe, they always talk about "good approximations" and Newton's law of gravitation rule! xD, (it can take you to the moon!).

11

As stated in other answers it is how much the gravitational force is different by on opposite sides of the earth that creates the tides. You can still show this using $a=\frac{GM}{d^2}$ but you need to consider the difference, not the absolute force on the earth. The sun while much more massive is just far enough away that it is getting to a much flatter ...

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