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Newton's original proof was in fact based on geometry (he hadn't invented calculus yet). Richard Feynman devised his own, simpler geometric proof for one of his famous lectures. You can find it in Feynman's Lost Lecture, by Goodstein & Goodstein, and in this article: Paths of the Planets from Hall & Higson. But since it's so much fun, I'll describe ...

8

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...

6

Suppose you pick two people at random. From one, you pluck a single hair from their head. Is it possible to tell who had the hair plucked by weighing the people? Technically, plucking a hair makes a person very slightly lighter, so you get a tiny bit of information about who had the hair plucked by weighing the people. But the information is very slight ...

4

Black holes this small will have very high Hawking temperature: $$T_H = \frac{\hbar c^3}{8 \pi G M k_B} \approx 10^{20}\,\text{K},$$ So, before this black hole can fall down even the diameter of an atom it will evaporate through Hawking radiation. As a result, the 1 tonne of black hole mass would be converted into the energy of very high energy particles ...

4

That's exactly the case. If you look at the trajectory of any given spacecraft, you will see that it has a few burns of the rocket engines punctuating very long periods just coasting along in orbit around some other body. For example, the flight path of Apollo 8 has something like eight different rocket burns: launch, translunar and transearth injection (to ...

3

The only way to do it is to put you temporarily in free fall. But as for the room you describe, I can only think of one type. Bring a scale with you next time you go down an elevator, and watch artificial gravity reduction at work! Heh heh.

3

The acceleration due to gravity changes not only on the surface of the Earth (depending on where you are) but also how high up you are (which varies by $\frac{1}{r^{2}}$, where $r$ is the distance from the center of the Earth to you). For more information on how it varies depending on location, maybe this will be of use? GOCE Delivers Best Gravity Map of ...

2

Partly because the magnitude of the gravitational force decreases as $\frac{1}{r^2}$, so as the distance from the center of the earth, $r$, increases, the magnitude decreases. The bigger reason for spacecraft is because they are constantly in free fall, and there is no way to feel gravity when you are falling freely. The spacecraft are falling and moving ...

2

I've never been at a theme park where you can mount into a plane at free fall. The photo that you posted is inside a reduced gravity aircraft. So you don't modify gravity, you are just falling.

2

why we always choose the center of gravity of the bicycle be the rotational center. We do not do that always, sometimes it is better to use the point in contact with the ground or some other point. We use center of mass when it leads to simpler equations than the other points. In problems dealing with torques or rotations we use the theorem T: the sum ...

2

A space traveler on an artificial satellite will be in freefall around the planet it is orbiting. So the ink will not experience any acceleration relative to the pen due to the planets gravity. On earth the ink gets sucked up by the pen due to capillary action, but is counteracted by gravity. While in orbit the full "force" of capillary action can be used ...

2

In the situation at hand, you'll never be able to achieve uniform circular motion. $\frac{d\vec{v}}{dt} = \frac{1}{m}\sum \vec{F}_{\text{ext}}$ This is a vectorial equation. If you look at the picture you've drawn, you have forces on the radial as well as the tangential direction. On the radial direction, there is the tension force, and $mg\cos\theta$. ...

2

I suspect that you may be under the mistaken impression that there is no gravity in space. This is a common belief since we all can see the astronauts floating in "zero g" when on the ISS are some other spacecraft. However, we can easily dispense with this misconception by asking "what keeps the ISS in orbit around the Earth if there is no gravity?". Of ...

2

As dmckee said in his comment, the black hole would fall towards the center of the Earth. To specifically answer this portion of your question: How dense would rock have to be to form a barrier? There is absolutely no density of rock or anything else that would stop or even slow it down. Even if you created this microscopic black hole on the surface ...

1

Both definitions are fine as long as you're careful with signs. Here's a derivation of your gravitational potential energy using the idea of the negative work done "by the field." (Note the initial negative sign.) U(r)=-W_\text{by field}=-\int\vec{F}_\text{field}\cdot d\vec{s}=-\int_{r=\infty}^{r=x}\underbrace{\frac{-GMm\,\hat{r}}{r^2}}_\text{Toward ...

1

If you only consider viscous dissipation within the droplet, this should indeed go to zero in the vanishing velocity limit: the (local) dissipation rate is quadratic in the velocity, so that decreasing the velocity by a factor of $\lambda$ reduces the (local and global) dissipation rate by $\lambda^2$. Of course, the process takes $\lambda$ times longer, ...

1

Most certainly it does: the variation can be measured by a sensitive acceleration called a Gravimeter (see Wikipedia page with this name) and is the basis for gathering data important for minerals exploration. Bodies of mineral ore distort the Earth's gravity and thus can be found by measuring the variation of the local gravitation as a function of position. ...

1

"Focus" is an inconvenient word if you're thinking of changing the potential, because if you do then the orbits are no longer conics and the word kind of loses its meaning. That aside, let me see if I understood your question correctly: Given a gravitational potential that's spherically symmetric around a central point $\mathbf{r}_0$, and which has a ...

1

I don't think that you really understand integration. Let me clear this up for you. In that question there is a rod of length l. You know how to calculate gravitational force between two point masses but not in continuous mass bodies. If you apply the formula to find the gravitational force you don't know what to take the distance as because it is ...

1

As is well-known from Newton's shell theorem, the gravitational field $g(r)=\frac{GM}{r^2}$ outside a spherically symmetric mass-distribution is the same as if the total mass $M$ sat in the center. It seems that OP wants to calculate the oblateness of Earth under the simplifying assumption that the backreaction (which the re-distributed mass has on Earth's ...

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