# Tag Info

7

While the constant itself is not dependent on anything else, its effect on the universe is dependent. For example, $G$ is a sort of ratio that relates the force between two objects at a given distance and of two certain masses. $G$ does not affect the electrical force between two charged particles - that's Coulomb's constant; it does not affect the time it ...

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 ...

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 ...

5

This question touches on the distinction between weight and mass which are used confusingly synonymously in many Western everyday-languages (not so in e.g. Russian). Weight is the force of gravity pulling you down. It is proportional to the object's mass which explains the confusing everyday usage of the word. A bathroom scale really measures the force ...

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.

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 ...

3

This is an interesting case of a kind of 'gravity spring' when the sphere is within a certain range of the torus. If we consider the torus as a ring and the sphere as a point mass by Newton's First Theorem, then we can show that for a ring with radius $a$, the force between the two objects decreases as they are brought closer together, over range of ...

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 ...

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.

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 ...

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 ...

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 ...

2

Take the orbit equation (using $u=\frac1r$): $$u(\theta)=A\left(1+e\cos\theta\right)$$ Where $A$ is some constant that you can work out from the differential equation if desired, and $e$ is the eccentricity of the orbit. Now, what happens when $\theta$ goes through an angle of $2\pi$? Notice that we would not have returned to the original radius if ...

2

No, it can't: free fall and zero gravity are the same thing. On that equivalence hangs the theory of General Relativity, in fact.

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 ...

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 ...

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

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 ...

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 ...

1

In physics we recognize two different kinds of mass: inertial and gravitational. Inertial mass tells us how much an object resists a change in motion - or how much force is needed to effect an acceleration. Gravitational mass describes how much attraction (due to gravity) an object experiences as a result of gravity. Now despite very careful experiments, ...

1

The equivalence principle applies only for a very small size lab, where it is not possible to make the parallel vs radial distinction you mentioned. Small size indicates, it is local. More over, there are other ways to distinguish between the two, if the size of the lab is big enough. One example - gravity changes with height and the acceleration inside an ...

1

The kgwt is not a unit which you should be using rather use the newton (N) as the unit of force. 9.8 crops up in a number of situations. The force of gravitational attraction on a mass of 1 kg on the Earth is 9.8 N. Another way of putting that is that the gravitational field strength on the surface of the Earth is 9.8 N/kg. The acceleration due to ...

1

Use that the gravitational force between two (point) masses $m_1, m_2$ separated by a distance $r$ depends on the product of the masses $m_1 \times m_2$ and is inversely proportional to the square of the separation, $\frac{1}{r^2}$. What you have to do is figure what happens if one of the masses goes from $m_1 \to 5 m_1$ and $r \to \frac{r}{6}$, so you don't ...

1

The formula you are looking for is the one for gravitational force $$F_g = \frac{Gm_1m_2}{r^2}$$ since $G$ and $m_2$ do not change in your example, we can write this more concisely as $$F_g \propto \frac{m_1}{r^2}$$ You will then find the answer by replacing $m_1$ with $5m_1$ and $r$ with $\frac{r}{6}$.

1

See this Figure For point mass $m_1=1$, and box mass $m_2=3$, and box size $a$.

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 ...

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. ...

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

@ 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 ...

1

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 ...

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