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I would not worry about air resistance for the balloon - being large and filled with water it will matter less than for the little ball. I would, however, worry about the strength of the catapult - how does its launch velocity change with mass of the object. If the catapult is made with an elastic that is pulled back by e certain amount then it van do a ...


-1

Today I found this on arxiv.org Shouryya Ray Paper It's the official paper from Shouryya Ray: An analytic solution to the equations of the motion of a point mass with quadratic resistance and generalizations Shouryya Ray · Jochen Frohlich why did he write it two years later from his original solution ?


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Naturally, I know it to be true that the moon goes around the Earth and that the Earth goes around the sun. While it definitely looks like it if you view it with an astronomy program, it is not entirely correct and a misconception. Sun, earth and moon are not nailed to a specific location, there is no axis stuck in vacuum. Each body attracts each ...


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No, you don't need to account for the mass of humans. We're a part of the total mass. Children who grow and gain weight are merely transferring mass from other parts of the Earth to them. The mass of humanity versus the mass of everything else on the Earth is a zero sum game. To see whether the Earth is changing mass, you need to look outside the box (or in ...


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Lots of places state that the Earth's gravity is stronger at the poles than the equator for two reasons: The centrifugal force cancels out the gravity minimally, more so at the equator than at the poles. The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field. TL;DR version: There are ...


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There is approximately 7 billion people on earth. If we take some average of about 75 kg each, that means $$ m_{\rm population}=7\cdot10^9\times75\,{\rm kg}=5.25\cdot10^{11}\,{\rm kg} $$ The mass of the earth is roughly $$ m_{\rm earth}=5.97\cdot10^{24}\,{\rm kg} $$ which is about 13 orders of magnitude bigger than the total population of the world. So no, ...


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Yes you need to include the mass of the 7 billion or so humans in the total mass of the Earth, though since their total mass is something like $5 \times 10^{11}$kg and the mass of the Earth is around $6 \times 10^{24}$kg humans make up only about 0.00000000001% of the total mass. We don't make any great contribution to the escape velocity. The point of your ...


2

The Earth/Moon orbit is not truly metastable. As someone alluded to, the moon is actually very very gradually getting further from the Earth. Conversely Mars' moon Phobos is gradually moving closer to Mars over time and Phobos will eventually crash into Mars.


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From the answer... ..my question I have learnt that Newton's 3rd law of motion is a direct consequence of law of conservation of energy. When a body moves in a certain direction and an opposing force acts on it , it exerts a reacting force (by Newton's law) ... Hence, the body loses its kinetic energy. Problem ....gravitational force: ...


1

My slight issue and where I think I might be missing something, is why doesn't the Earth's speed just mean it can just fly away from the moon and just leave it flying in it's tangential velocity? Even though you have drawn the force of gravity, you are not thinking about it. First of all the force in general is bidirectional, the earth pulls the moon ...


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You might as well imagine that it is fixed. Basically, as far as the moon is concerned gravitationally, only the Earth exists. For the Earth; our sun. For the Sun; a black hole, ect... and everyone of them has an inherited orbital velocity from their 'parent'. Orbit is not a place. As for why the moon doesn't crash, it formed outside of the Earth's Roche ...


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Let me try this way: the Sun isn't only pulling on the Earth, it's pulling on the Moon as well. The pull on the Earth is almost the same as the pull on the Moon, so the net effect of the Sun on the relative motion between the Earth and Moon is very small. Recall Galileo's law of motion: if you drop two objects close together from the same height, they ...


3

is why doesn't the Earth's speed just mean it can just fly away from the moon The moon and the Earth fall towards each other due to their mutual gravitation. The Earth doesn't have enough speed, relative to the Moon, to 'just fly away'. Equivalently, the Moon doesn't have enough speed, relative to the Earth to 'just fly away'. Here's an image I ...


0

Here's a simple argument that doesn't require any knowledge of fancy stuff like equipotentials or rotating frames of reference. Imagine that we could gradually spin the earth faster and faster. Eventually it would fly apart. At the moment when it started to fly apart, what would be happening would be that the portions of the earth at the equator would at ...


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The point is that if we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface,$^1$ see e.g. this Phys.SE post. Now, because the polar radius is smaller than the equatorial radius, the density of equipotential surfaces at the poles must be bigger than at the equator. Or equivalently, the field strength$^2$ $g$ ...


1

All celestial bodies loose atmosphere due to a portion of the gas "near space" exceeding escape velocity. The velocity distribution of an ideal gas can be found using the Maxwell-Boltzmann distribution. So an easy approximation for this problem is to say we only want $10^{-6}$ of the molecules to have escape velocity. Using oxygen at 300K, results in an ...


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If you take something like Neptune and pass it through Earth's orbit perpendicular to the ecliptic so it collides directly with the Earth-Moon system at, lets say, 0.1c you would remove the Earth rather quickly. Get another if you want to disappear the moon as well. Neptune isn't so big that it would disturb everything else, maybe a wobble in Venus or Mars ...


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Paily's answer is the correct one. Mechanical energy is defined as the sum of kinetic and potential energy and equal to the work done by non-conservative forces. When these are absent (as when they act perpendicular to displacement) they do no work, therefore ΔK+ΔU = 0 and K+U = constant. In short, the principle of conservation of mechanical energy: any ...


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You hold a stone and let it go. What work will the Earths gravity do on the stone to bring it down to the ground? This much: $$W=\Delta K=K_2-K_1=K_2$$ Let's remember this for now and do something different. How much potential energy is lost during this fall? This much (energy conservation): $$E_{\text{total }1}=E_{\text{total }2}$$ $$U_1+K_1=U_2+K_2$$ ...


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Here are a few points to keep in mind: Potential energy is always described as the potential energy of the system. For example, the gravitational potential energy of the Earth-Moon system, belongs to the system as a whole, not the Earth or the Moon individually. So for your example, if you are for instance throwing a brick upwards, it would be the ...


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If I understand you correctly, your mistake is in using friction as an analog to gravity. Because friction is a non-conservative force the work done is dependent on the path taken. Furthermore the energy "lost" due to friction is stored in a way that is not spontaneously reversible within the system (e.g. heat, plastic deformations, etc.). Gravity on the ...


0

Nothing is actually stored. (You will not find anything "in" the body :) ) The increase of potential energy means in this case that there is a force (of gravity) acting on a body, and the body's movement away from the source of this force increases the distance the body can p o t e n t i a l l y travel under the influence of this force. So if the body is ...


3

Your wife could be right about the rocket. Whenever we say something is small physically, we need to be sure what it is small with respect to. In the case of a thread, the drag force exerted by air beats the centripetal force keeping it taut. Because centripetal forces scale with mass, a denser thread will work. The issue facing a space elevator is ...


1

A space elevator would need to extend well beyond the altitude of geosynchronous orbit. The center of gravity (not center of mass) of the space elevator system would need to be at geosynchronous altitude. The most common architecture is to attach the outer end of the elevator to a counterweight. If the cable broke, this counterweight would orbit ...


3

Yes, the gradient in spherical coordinates contains angular terms. It has to, because the force need not point to the center. The relevant equation is $$\Delta U=\frac {\partial U}{\partial r}e_r+\frac {\partial U}{r\partial \theta}e_\theta+\frac {\partial U}{ r \sin \theta \partial\phi}e_\phi$$ You have no $\phi$ variation, so can ignore the last term.


2

All finite-sized objects act like point masses at sufficiently large distances. A finite-sized three dimensional object of mass $M$ has a unique minimal bounding sphere with some radius $r$ and center $c$. For any distance $R>r$ from the center, the magnitude of the gravitational force on a small test body of mass $m$ is bounded by $\frac {GMm}{(R+r)^2} ...


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This is certainly not the absolute truth, since every human on this planet is a living witness to the fact that Earth is not pointlike: It has finite volume. However, there's a very good mathematical reason why we can consider the sun and Earth as pointlike when doing calculations in celestial mechanics: It's because the gravitational field due to a ...


1

Point masses are merely an approximation of the motion of an extended body by the motion of its center of mass. It's a good approximation for small, mostly homogeneous (at least spherically layered) bodies, e.g. planets moving around the sun. But even for the orbital motion of the Moon around the Earth it's not a good approximation anymore.


2

Excluding F, G, M, and m (you've already used those names in this expression), you could label that distance any letter from a to z or from A to Z or from $\alpha$ to $\omega$. Or whatever. It doesn't matter. It's a variable. That said, there are conventions. It's best not to call that distance v, for example. The symbol v usually means a velocity or speed, ...


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Here are couple of references that describe professional uses of a post-Newtonian formalism to model the planets and the Earth's Moon: Standish, et al. "Orbital ephemerides of the Sun, Moon, and planets," Explanatory Supplement to the Astronomical Almanac (1992): 279-323. The relevant equation is 8-1 on page 3. Petit and Luzum (eds.), "IERS Technical Note ...


3

Small elaboration on t.c's answer: If you say that $$F = \frac{G\ M_e}{R_e^2}m = g\ m$$ at the surface of the earth $R_e$, then at other heights you can write (for $h << R_e$): $$\begin{align} \\ F &= \frac{G\ M_e}{(R_e+h)^2}m\\ &= \frac{G\ M_e}{R_e^2(1+\frac{h}{R_e})^2}m\\ &\approx\frac{G\ M_e}{R_e^2}(1-2\frac{h}{R_e})m\tag1\\ &= g ...


1

It depends what you consider a cube or a sphere. Is the Earth a sphere or do the mountains make it something else? As you do seem to understand more massive planets have more gravity, which shapes them into a sphere. This is because matter on the outside of the planet is attracted to the center. On Earth a loose rock is more likely to fall from the top of a ...


2

You can use the Newton's law of gravitation which is given by: $$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$ The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so: $$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822 = g$$ You can adjust $r^2$ ...


1

Newton very much relied on the works of his predecessors, and sometimes his contemporaries. This is what scientists do do, and should do. Newton did not "figure out" his universal law of gravitation completely on his own. He instead expanded upon the works of Galileo, Kepler, Halley, Huygens, Hooke, and others. There are issues with regard to priority for ...


3

Although the story of the apple is most likely made up, Newton did try to find a connection between the motion of falling objects, the Moon, and the planets. He started with the acceleration of an object moving at constant speed $v$ in a circle. Within a small time interval $\Delta t$ the object sweeps out an angle $\Delta \theta$. The change in direction of ...


2

So the acceleration is independent of the mass. So in this question both the bodies should fall at the same time. Your answer is correct if you dropped those objects on the Moon. However, the Moon has no atmosphere. You are ignoring drag, or "air resistance". Suppose you drop a ping pong ball and a super ball of the same size as the ping pong ball. Both ...


1

Consider the force applied by air resistance $F_a$, equal for both objects. Then the equations of motion are: $$F_a - \frac{GM_\oplus M}{r^2} = Ma$$ and $$F_a - \frac{GM_\oplus m}{r^2} = ma$$ Solving for $a$ in each case, you'll find that the acceleration due to gravity is the same in both cases (as expected), but the acceleration due to $F_a$ is different ...


3

I suspect it's all in what they meant by "air resistance is the same ." For example, if both bodies are the same size and shape, the equation for air resistance as a func of velocity is the same, but the greater mass will be less affected by this equal force magnitude. Air resistance being "same" is not the same thing as being "neglectable" or "zero."


3

Well, there are 4 parts to the right-hand side; let's look at each in turn. The first $M$ is the mass of the gravitating body. You would expect that the force it exerts should be larger if it has a larger mass. Moreover, it's not unreasonable to think the force should be directly proportional to this mass: twice as much mass should grab things with twice as ...


0

r is the distance between the two masses v is the relative velocity a is the relative semimajor axis* * Two bodies orbiting each other trace out two separate ellipses in an inertial frame. The smaller body traces a larger ellipse, and vice versa. The relative semimajor axis (a) is equal to the sum of the semimajor axes of these two ellipses. The relative ...


0

For a satellite in a nonperturbed orbit, the change in gravitational potential energy does equal the change in kinetic energy. Increasing the size of a circular orbit requires energy. Half of the energy is used to lift the satellite, while the other half is used to speed up the satellite. For a satellite in a circular orbit perturbed by atmospheric drag, ...


2

It's known that the general solution of the Laplace equation $\nabla^2\Phi = 0$ in spherical coordinates with azimutal symmetry is given by: $$\Phi(r,\theta)=\sum_{l=0}^\infty\left(A_l r^l + B_lr^{-(l+1)}\right)P_l(\cos \theta),$$ where $A_l$ and $B_l$ are given by boundary conditions. We also have that the function $\dfrac{1}{|\mathbf{x}-\mathbf{x'}|}$, ...


3

Newton law of gravitation is given by: $$F = G \frac{m_1 m_2}{r^2}=\left(G \frac{m_1}{r^2}\right) m_2$$ The gravitational constant, $G$, the weight of Earth, $m_1$, and the radius are constants, so: $$G \frac {m_1}{r^2}=(6.6742 \times 10^{-11}) \frac{5.9736 \times 10^{24}}{(6.37101 \times 10^6)^2}=9.822$$ Hence, the equation simplifies to $$F =(9.822) ...


1

See the wiki page . There is no difference between the two . Weight of a body of mass $M$ is $M.g$ which is equal to gravitational force on the body.


0

If you exclude the forces of air resistance then this becomes the classic "Which falls faster? The baseball, or the cannonball?" In the case of the cannonball and the baseball, the cannonball will have negligibly more force applied to it, but will accelerate more slowly than the baseball. In the end, they will both hit the ground at the same time. Big ...


2

Yes, adding mass to a toy car should at least in principle make it accelerate down a ramp faster. The total force on the car is in the "forward" direction, with magnitude $$F=m g \sin\theta\ -\ m g C_{rr}\cos\theta\ - \tfrac12 \rho v^2 C_D A\ ,$$ where $m$ is the car's mass, $g$ is the acceleration due to gravity at Earth's surface, $\theta$ is the angle ...


3

While the speed and the position of an object change gradually (because of $x(t) = \int_0^t v(t') dt'$ and $v(t) = \int_0^t a(t') dt'$) the acceleration can change instantly therefore it doesn't matter for the acting force wether you just dropped the stone or not. So while you're holding the stone it feels exactly the same acceleration as you do (the force ...


4

Acceleration only happens if a force is acting on an object. When you are holding the stone, the force which is accelerating the elevator is transmitted to the stone through your grip and you, the elevator and the stone form a single rigid collection of objects all accelerating at g/2. When you let go the only force left acting on the stone is gravity and ...


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While the stone is still travelling on the elevator, there are two forces acting on it, the force from the elevator to the stone, as well as the weight due to gravity. The moment the stone leaves the elevator, it becomes a free falling object. The elevator stops giving a force to the stone, and the only force remaining is its weight due to gravity. ...


2

The space elevator probably deserves an entire series of questions (and I am sure there have been plenty of posts), but if we stick to this particular version, there are a couple of problems with it. First of all, a space elevator needs a counterweight in an orbit that is higher than the areostationary orbit (the Martian equivalent of Earth's geostationary ...



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