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43

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$m g = k v^2$$ And thus $$v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...

39

Intuitive explanation Suppose you drill two, perpendicular holes through the center of the Earth. You drop an object through one, then drop an object through the other at precisely the time the first object passes through the center. What you have now are two objects oscillating in just one dimension, but they do so in quadrature. That is, if we were to ...

19

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...

18

The simple answer to Why does the Pluto's orbit crosses the Neptune's orbit is to just say that's the way it is. For any object orbiting in a central inverse square law field, like the gravitational field of the Sun, the stable orbits are ellipses with the Sun at one focus. The ellipses can be almost circular like the Earth's orbit or wildly eccentric like ...

12

Background You would need a very sensitive instrument to measure the daytime vs nighttime difference in g. It is not 0.006 m/s2. It is much, much smaller than that, about $6\times10^{-11}$ m/s^2. Your 0.006 m/s2 is the gravitational acceleration toward the Sun at distance of 1 AU. The Earth as a whole is accelerating sunward at 0.006 m/s2. You cannot ...

11

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...

8

Phil's answer, while beautifully illustrated, is a little incomplete. It relies on the fact that in the case of the tunnel you're solving the one dimensional projection of the low earth orbit satellite, but doesn't prove this. I do this below. The force applied on the object, for a sphere of uniform density, is actually : \begin{eqnarray} F &=& - ...

6

A couple of decades ago, I put quite a bit of time into building a high-precision differential gravimetry setup using a pendulum. It was a lot more difficult than I'd expected, and I was never very successful. If you want to do differential gravimetry, it would probably be much more doable to measure differences between different heights above the earth's ...

6

Pluto's average distance from the Sun is larger than Neptune's but Pluto's orbit has a higher eccentricity – the elliptic orbit is more squeezed, less uniformly circular, and such ellipses simply do intersect each other. The elliptical orbits with properties first identified by Kepler's laws do follow from Newton's laws of gravity. I stress that the ...

5

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse. A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is ...

5

No, the entire weight will not directly rest on the base of the slanted container (although it does indirectly). There are a number of ways to approach this, but the easiest way is to observe that the total force acting on the bottom of the container is equal to the sum of the hydrostatic pressure force (the pressure at the bottom of the container multiplied ...

5

I would say, that your two first statements are not wrong, i.e. you can defend them in a discussion if you explain what you mean by extension and special case. Your conclusion, as you suspected anyway, is wrong however. Special relativity and Newtons law of gravitation are not the same thing. In fact they deal with different aspects of a physical theory. ...

4

In a completely ideal world, where air resistance was not present, the bomb would continue to move forward and the same horizontal speed as the airplane it was dropped from. Gravity only acts in the perpendicular direction, thus has no effect on the horizontal component of the bomb's velocity. In reality though, air friction reduced the speed of the bomb as ...

4

It never becomes impossible per se, but at some point there could be so much gravity that construction of a working rocket would be beyond our current ability to engineer something that could work. That is, it might take impracticably huge quantities of fuel, or require materials stronger than we can construct. There are just a couple of amazingly simple ...

4

Following the same orbit at greater velocity would mean that the satellite has greater acceleration (its velocity is changing at a greater rate than if it follows the same route more slowly). But the only force acting on it is earth's gravity, which creates a particular acceleration at any given altitude above the earth. It can't go any faster or slower ...

4

Let $\mathbf x(t)$ be the path of a particle. Let $\mathbf F(t)$ be a force acting on the particle as a function of time, then the work done by the force from time $t_a$ to time $t_b$ is \begin{align} W(t_b, t_a) = \int_{t_a}^{t_b} \mathbf F(t)\cdot \frac{d\mathbf x}{dt}(t)\, dt. \end{align} where the center dot denotes dot product; \begin{align} ...

4

Yes, that can happen. It is somewhat realized in positronium, a bound electronic state where an electron and a positron revolve around each other. Both have the same mass, so they could (classically) have the same spherical orbit. With Newton, you have an attractive force for two equal bodies of mass $m$ of $$F = G \frac{m^2}{d^2}.$$ The centripetal ...

4

The faster you go, the less velocity you theoretically can gain from a gravity assist. The reason for this is that the faster you go the harder it is to bend the orbit. To proof this we have to use the patched conics approximation, which means that while within a sphere Kepler orbits can be used. The sphere can be simplified to be infinitely big, since the ...

4

The question refers to "laboratory" measurements, i.e., local ones. Such a measurement can only be sensitive to the gravitational acceleration of a test mass relative to the laboratory. For example, if one drops a mass in a vacuum column and measures the time it takes to hit the floor, the acceleration inferred is the acceleration of the mass relative to the ...

3

You can easily convince yourself that this is not true. Just imagine a test particle on one side of a spherical shell. If the density is uniform, the theorem will apply. Now, however, decrease the density of that half of the shell until it's 0. Obviously this other half would attract the test particle. Alternatively, if changing density is not an option and ...

3

Assuming that $m_1$ and $m_2$ take up a finite amount of space (e.g., two spheres of mass with radius $r_0$), that equation isn't even valid for $r < r_0$, so there's no inconsistency. The derivation follows from Gauss' law; it is analogous to the application of Gauss' law in electrostatics; the $m_1$ and $m_2$ are the mass enclosed at some distance $r$. ...

3

I would say Newtonian gravity is the limit of general relativity as gravitation becomes weak. See it this way: the limit when the temperature-amplitude is small of metal extension can be written as: $$L = L_0 (1+\alpha \delta t)$$ If $\delta t$ is quite large this equation is no longer applied. Special Relativity doesn't care about gravity, like the ...

3

You're confused. The number you're referring to, $6.6738410\times 10^{-11}\ \text{m}^3\ \text{kg}^{-1}\ \text{s}^{-2}$, is Newton's gravitational constant, $G$. This is not a formula or equation. The 'variables' are actually units: meters (m), kilograms (kg), and seconds (s). The formula where this constant most commonly appears is for Newtonian gravity: $$... 3 For a uniform circular orbit of radius r, the acceleration is$$\tag{A} a~=~ \omega^2r, \qquad \omega~=~\frac{2\pi}{T},$$where T is the orbital period. Comparing eq. (A) with Kepler's third law$$\tag{B} T^2 ~\propto~ r^3,$$we conclude that the gravitational acceleration$$\tag{C} a~\propto~ r^{-2} $$is proportional to the inverse square ... 3 One can get an order of magnitude estimate of the maximum speed attainable by gravitational slingshots without doing any real calculation. The 'rough physics' reasoning goes as follows: The gravitational field of the planets used for slingshots needs to be strong enough to "grab" the speeding spaceship. As a planet cannot "grab" a spaceships moving faster ... 2 The proof is most easiest if we use the vector notation. We have$$\vec \tau = \int {d\vec \tau } = \int {(\vec r \times dm\vec g)} = \left( {\int {\vec r dm} } \right) \times \vec g$$where I have used the assumption that near the earth \vec g is constant. Now according to the definition of center of mass we have,$${{\vec r}_{cm}} = \frac{1}{M}\int ...

2

The symbols $\text{m}$, $\text{kg}$, $\text{s}$ aren't variables that depend on a particular situation. They are the units. You can write $G = 6.67384\times10^{-11} \frac{\text{m}^3}{\text{kg}\cdot\text{s}^2}$, which means that $G$ is measured in cubic meters per kilogram-squared second, exactly the same way that you can say the Earth's radius is $6370 ... 2 The formula: $$\Delta U = mgh$$ is an approximation that applies when the distance$h$is small enough that changes in$g$can be ignored. As you say, the expression for$U$is: $$U= -\frac{G M m}{r}$$ So the change when moving a distance$hupwards is: $$\Delta U = \frac{GMm}{r} - \frac{GMm}{r + h}$$ We rearrange this to get: \begin{align} ... 2 Your first potential energy arises from the approximation that the graviational field is approximately constant for "small heights" , i.e.\frac{GMm}{r^2} \approx mg$The full law leads to your second formula, the approximation to the first. For heights above the earth, it is justified, as we can see by taylor expanding$\frac{1}{r^2}\$ around the ...

2

No. the whole weight will not act on the base of the container 2. If the whole weight had acted on the base of container 2, then the pressure on the base of container 2 would be equal to that of container 1 i.e., mg/A, where mg is the whole weight of the fluid and A is area of the base. But as you know the pressure at a depth depends on the height of the ...

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