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69

It sounds like you are imagining that what satellites do is go up through the atmosphere, break though into outer space, and hang there. That is not right. If you simply go straight up to outer space (say 300 km above Earth's surface), gravity will pull you right back down, even if you've left the atmosphere, and you'll crash back into the Earth. Gravity is ...


47

I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like: The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity. To write down the equation of motion ...


30

I highly recommend you download Kerbal Space Program and see for yourself (there's a free demo version)! Typically the goal of a satellite is to orbit, and thus as the other answers address, you must build significant horizontal velocity. Indeed if the Earth didn't have an atmosphere, you could orbit a few km above the surface, so the main goal is building ...


25

If the mass/charge is symmetrically distributed on your sphere, there is no force acting on you, anywhere within the sphere. This is because every force originating from some part of the sphere will be canceled by another part. Like you said, if you move towards on side, the gravitational pull of that side will become stronger, but then there will also be ...


24

The force you can exert is your mass times your acceleration. By the equivalence principle, just standing still is equivalent to accelerating at 9.8 m/s2, which is where the force of your weight comes from when you just stand still. But it is easy to accelerate more - like when you jump. The force is only limited by your ability to push yourself off ...


21

Yes, this is possible. It is perfectly fine for a mass configuration to produce, for points outside a sphere of radius $R$ centred at $\mathbf r_0$, a gravitational field identical to that of a point mass at $\mathbf r_0$, and still be completely empty inside a smaller sphere of radius $a$ around $\mathbf r_0$. The spherical-shell model you describe is ...


19

The answer is that inside a spherically symmetric shell of matter (your hollow earth or massive beach ball) there is no gravitational force anywhere - you will not "fall" in any direction, whether you are at the centre or not, regardless of the radius of the sphere. This is a classic result of both Newtonian Gravity, and Einstein's General Theory of ...


7

I assume that you're imagining two sticks whose bases stay stuck still to the ground as they tip over. In this case, what we should do to calculate the tipping rate (I wouldn't exactly say "falling"-- I think it's misleading) is consider the torque applied to each stick as it tips. What the answer really comes down to is the struggle between torque (the ...


6

Emilio Pisanty's answer is a great one to this particular answer, i.e. there is no in principle bar from the laws of physics to a structure like yours and it would have the zero gravity inside property that Emilio describes. But, from a materials perspective, it would be pretty much impossible for a planet like this to form unless very small. The stablest ...


6

Use a lever. For an application like driving branches in the great outdoors, you would need to come prepared, or locate the site next to something heavy. Anchor one end of the lever under something heavy. Attach something to the branch to press against, and put the branch between you, at one end, and the fulcrum at the other. The lever acts as a force ...


5

If you think about Newton's third law and standing still vs jumping. When you stand still the ground exerts a reaction force on you which is equal and opposite to your weight by Newton's third law. If you jump upwards at the point where you begin to drive upwards you are applying a greater force on the ground than the standing still case, this difference is ...


4

Rockets take the shortest path to reach their orbit. If all we want to do is pop up to LEO and come back down again, then we do go straight up. See the first two Mercury missions for an example - they landed north of Nassau. If you want to end up in orbit, you need substantial horizontal speed. Turning at right angles is the least fuel-efficient way to do ...


4

According to the shell theorem, the gravitational force inside such a hollow spherical planet would be 0 On the outside, gravitational force would be as if the planet was a regular planet. At some point during your fall, the gravitational force pulling you to the center would decrease to 0. If there's atmosphere inside the planet, it would slow you down ...


4

Because of a convention wherein zero gravitational potential is said to be at infinity. See Wikipedia: $V(x) = \frac{W}{m} = \frac{1}{m} \int\limits_{\infty}^{x} F \ dx = \frac{1}{m} \int\limits_{\infty}^{x} \frac{G m M}{x^2} dx = -\frac{G M}{x}$ "By convention, it is always negative where it is defined, and as x tends to infinity, it approaches zero." ...


4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...


3

In your context, the second interpretation is correct. The fact is that falling objects accelerate both on Earth and on the Moon. The sentence is saying that the amount of this acceleration, regardless the source, is six times greater on Earth than on the Moon. In other words, things accelerate towards the surface of Earth six times faster than they ...


3

We'll start with your second question, as that's (much) simpler. Close to the surface of the earth, it's safe to assume that the force of gravity is proportional to the mass of your object: $$F_G = mg$$ where $m$ is the mass, $F_G$ is the force of gravity, and $g$ is a constant (for the earth $g \approx 9.8 m/s^2$.) Then Newton's second law tells us that ...


3

Cosmic Velocity has nothing to do with infinity. A cosmic velocity is the minimum speed directed in the necessary direction to escape the gravitational attraction of a cosmic body such as a planet, a star, or a galaxy. Here is a paper which a student wrote about the four cosmic velocities. I don't know if his exact classifications are in common usage, but ...


3

The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ...


3

I'm assuming you haven't taken any physics courses, so let's start by explaining the concept of a force. Forces are the central focus of classical mechanics. Basically, a force is a push or pull on an object as a result of its interaction with another object. When applied to an object with mass, a force causes the object's velocity to change in some way. ...


3

The rotation period $T$ is given by $$T=2\pi \sqrt{\dfrac{a^3}{G(M_\text{Sun}+M_\text{planet})}}$$ where $a$ is the sum of the half axes of the ellipse. Routhly: $M_\text{Sun}=2\times 10^{30}$ kg $M_\text{Earth}=6\times 10^{24}$ kg $M_\text{Jupiter}=2\times 10^{27}$ kg If you assume both Earth and Jupiter are orbiting around the Sun (and neglect the ...


3

I was wondering that if they were orbiting in same orbit then will they both have same time period? If yes, then why because as they both have different angular momentum and both have so much of differences. I'll break this down into two parts, first looking at the period of individual objects orbiting the Sun at a distance of one astronomical unit (but ...


2

I'm sure this is an incomplete answer, and would only add to those already excellent, more knowledgeable answers, but I do know that rocket scientists try to situate their launch site as close to the equator as they can, so the Earth's diameter gives some extra "throw" as compared to higher latitudes. I expect when the rocket tilts toward the horizontal, it ...


2

Yes, you can. The amount of force you can exert on an object is limited only by the geometry and strength of your muscles. However, Newton's 3rd law dictates that however much force you exert on an object, the object will exert the same amount of force on you, in the opposite direction. So, if you exert a force larger than your weight down on a stick, ...


2

If there is no resistance then there will be no net torque applied about the center of mass. So any initial rotational speed will remain. The rotation center is going to be the center of mass. The effect of gravity will be to accelerate the center of mass, and it will have no effect on the rotational motion of the body. See this accepted answer for a ...


2

Yes it is theoretically possible, as discussed in the other answers and indeed we already do a variation of harvesting a planet's orbital kinetic energy in the space navigation manoeuvre called "gravity assist" or "slingshot" to boost a spacecraft's speed without expending propellant. Here one makes one's spacecraft "collide" with a planet (i.e. make a very ...


2

As per @lemon mentioned : You can treat the sphere as a single point (this follows from point 1 of the shell theorem). So you just need to integrate along the length of the ring. So I just did like that only. Taking sphere as point mass then point mass will experiences one gravitational force of attraction that is: $$df=cos\theta$$ As vertical ...


2

Gravitational force is really weak compared to the other fundamental forces, so it's very difficult to measure the gravitational constant. This is how Cavendish did it without knowing the Earth's mass: He put two lead balls on either end of a long bar. He hung the bar at its center from a long twisted wire with known torque. Then, he placed two really ...


2

This, in fact, is not too different from the flat surface case, the difference lies only in the distribution of the force components along the '$x$ and $y$' components of velocity. In the flat surface projection case, you have acceleration only along the vertical direction. The given case can be converted to a non-inclined plane case by mentally rotating ...


2

In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific ...



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