# Tag Info

85

Using your definition of "falling," heavier objects do fall faster, and here's one way to justify it: consider the situation in the frame of reference of the center of mass of the two-body system (CM of the Earth and whatever you're dropping on it, for example). Each object exerts a force on the other of $$F = \frac{G m_1 m_2}{r^2}$$ where $r = x_2 - x_1$ ...

46

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 ...

42

No. All parachutes, whether they are drag-only (round) or airfoil (rectangular) will sink. Some airflow is needed to stay inflated, and that airflow comes from the steady descent. Whether your net descent rate is positive or negative is a different question. It is quite easy to be under a parachute and end up rising (I have done it myself), you just need an ...

35

As other answers say, if someone just jumps off of the international space station(ISS), they would still be in orbit around the earth since the ISS is traveling at 17,000 miles per hour (at an altitude of 258 miles). Instead of just jumping, imagine the astronaut had a jet pack that could cancel that speed of 17,000 miles per hour in a very short time ...

25

As a very rude guess, fresh snow (see page vi) can have a density of $0.3 g/cm^3$ and be compressed all the way to about the density of ice, $0.9 g/cm^3$. Under perfect conditions you could see a 13 feet uniform deceleration when landing in 20 feet of snow, or about 4 meters. Going from $30 m/s$ to $0m/s$ (as @Sean suggested in comments), you'd have ...

24

It would be possible in theory, but only in a very side-thinking way: if you make a parachute so large it encapsulates the whole Earth, it will in effect act as a balloon and not fall down, due to the internal pressure of the atmosphere. This wouldn't work in practice for obvious reasons, but maybe in Kerbal you might be able to do something like it..

21

While everyone agrees that jumping in a falling elevator doesn't help much, I think it is very instructive to do the calculation. General Remarks The general nature of the problem is the following: while jumping, the human injects muscle energy into the system. Of course, the human doesn't want to gain even more energy himself, instead he hopes to transfer ...

20

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 ...

15

This is another chance to use one of my favorite approximations ever! I first offered it as an answer to a question about how deep a platform diver will go into the water. Now is the chance to use it again! Issac Newton developed an expression for the ballistic impact depth of a body into a material. The original idea was expressed for materials of ...

14

@Señor O gives a very good answer, but he assumes an ideal deceleration. Based on a viewing of the scene, Anna sinks a little under a meter, while Kristoff doesn't sink more than half a meter. Since they fell about 200 feet (about 60 m), my initial estimate for their impact velocity is (assuming no air resistance): $v = \sqrt{2gh} = \sqrt{2*60*9.8} ... 13 Analyzing the acceleration of the center of mass of the system might be the easiest way to go since we could avoid worrying about internal interactions. Let's use Newton's second law:$\sum F=N-Mg=Ma_\text{cm}$, where$M$is the total mass of the hourglass enclosure and sand,$N$is what you read on the scale (normal force), and$a_\text{cm}$is the center ... 12 it is because the Force at work here (gravity) is also dependent on the mass gravity acts on a body with mass m with $$F = mg$$ you will plug this in to $$F=ma$$ and you get $$ma = mg$$ $$a = g$$ and this is true for all bodies no matter what the mass is. Since they are accelerated the same and start with the same initial conditions (at rest and ... 12 It depends on how you define the problem. Humans have re-entered the atmosphere from the International Space Station many times, by riding in either a Space Shuttle or a Soyuz capsule. Someone re-entering without a spacecraft of some sort would obviously have to wear some kind of pressure suit (as Felix Baumgartner did in his jump). How elaborate is the ... 12 The paradox appears because the "rest frame" of the Earth is not an inertial reference frame, it is accelerating. Keep yourself in the CM reference frame and, at least for two bodies, there is no paradox. Given an Earth of mass M, a body of mass$m_i$will fall towards the center of mass$x_{CM}=(M x_M + m_i x_i)/(M+m_i)$with an acceleration ... 12 A parachute is a device specifically designed to create viscous friction. Viscous friction generates a force that: is oriented opposite to the velocity; is proportional to (a certain power of [*]) the velocity. So the falling velocity will increase until the drag force (pointing upwards) becomes equal to the weight of the falling object (pointing ... 11 He "only" flew at the maximum speed of 370 m/s or so which is much less than the speed of the meteoroids – the latter hit the Earth by speeds between 11,000 and 70,000 m/s. So he was about 2 orders of magnitude slower. The friction is correspondingly lower for Baumgartner. Note that even if he jumped from "infinity", he would only reach the escape velocity ... 11 You will die. Terminal velocity is a bit more than 50 m/s. The bottom of your ramp appears to have a radius less than 2m. That means you'll be exposed to more than 125g as you zip around the bottom. Nice knowing you. 11 As an addition to already posted answers and while realising that experiments on Mythbusters don't really have the required rigour of physics experiments, the Mythbusters have tested this theory and concluded that: The jumping power of a human being cannot cancel out the falling velocity of the elevator. The best speculative advice from an elevator ... 11 Other answers & comments cover the difference in acceleration due to drag, 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, the ... 11 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. ... 10 That is an excellent example for a nice quote I read on the internet: "Common sense may be common, but it certainly isn't sense" :-) As it is hard to lift heavy objects, we assume that it must be easier for them to drop. Now, Newton's laws point out that light and heavy objects will fall with the same velocity. But is there an intuitive reason? Yes! The ... 10 The reason that jumping can make a relatively large difference is that the kinetic energy is proportional to the square of the velocity. Thus relatively small changes to the velocity can result in relatively large changes to the kinetic energy. In addition, the velocity which a human can achieve in jumping is a substantial percentage of the velocity of fatal ... 10 If we are throwing two objects directly to the ground you are right. So from our kinematic equations: $$V_f = V_i + at$$ I would ask your teacher. What happens to the$V_f$if$V_i=0$? Then Follow it up with what would$V_f$be if$V_i$was very large? The initial velocity DOES have an effect here. HOWEVER: Make sure that you are not misinterpreting ... 9 Newton's gravitational force is proportional to the mass of a body,$F=\frac{GM}{R^2}\times m$, where in the case you're thinking about$M$is the mass of the earth,$R$is the radius of the earth, and$G$is Newton's gravitational constant. Consequently, the acceleration is$a=\frac{F}{m}=\frac{GM}{R^2}$, which is independent of the mass of the object. ... 9 indeed there would be a (very small) and homogenous pressure within the blob, coming from surface tension. This pressure is calculated by the Kelvin Equation and is significant in small droplets (reason for small droplets to have a higher vapour pressure than bulk liquid) In Your 100 m blob, this extra pressure is negligible of course. There is another ... 9 Nice theoretical answers (I can certainly appreciate them, I'm a mathematician). But why delve into theory when experiment is available? In this video you can see a skier jump from more than 200 feet and get head first into the snow, without a helmet. The video starts with the aftermath, if you want to see the jump right away fast forward to about 1 ... 8 In the global, cartoon, sense, yes, this problem is equivalent to having a whole row of carefully designed, placed and arranged ramps so that you fall onto the first one, get "flung" out such that you then land on the next one and so on, until dissipation wastes away the energy. Obviously this can be done since it is the same principle as is used in say ... 8 OK, based on the comments I interpret the question as Universal gravitation tells us that the gravitation force on a heavy object is larger than that on a light object, so why doesn't the heavy one fall faster? Start with Newton's gravitation (as simplified for objects in the neighborhood of the Earth's surface): $$F_g = mg$$ The answer arises from ... 7 Since from your previous questions you're obviously interested in learning how this is done I'll go into the detail of the calculation. Note that a lot of what follows can be found in existing answers, but I'll tailor this answer specifically at you. Time dilation is calculated by calculating the proper time change,$d\tau$, using the expression:$\$ ...

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