# Tag Info

5

Consider three bodies $Earth$, $B$ and $C$. Now, we take two cases. In the first case, we will observe gravitational attraction between body Earth and the heavier body ($B$); and in the second case, we will observe the gravitational attraction between body Earth and the lighter body ($C$). Case 1 Consider body $Earth$ and $B$, separated by a distance of ...

0

If you assume that the earth is immobile, the fall time if the same. If you consider that the earth moves because of the mass of the item, yes, the heavy item fall time is shorter (infinitesimal difference). If both items are dropped at the same time, they will hit the floor at the same time though. (This question has already be answered : Don't heavier ...

2

Clearly missed the point in this statement. For both the 1000kg and 1kg masses, the product of them both with the earth's mass is clearly the earth's mass, so F will be virtually the same for them both. The product is definitely not the same and the force on 1000kg ball is exactly 1000 times greater that the force on 1kg ball and much much more on the ...

11

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 ...

0

Wether the car will land on its 4 wheels or not is subjective to the cars centre of mass and the height of the cliff. As soon as 2 of the car's wheel are of the cliff , the car is not in equilibrium. Therefore the car will tilt toward the front and then depending on its centre of mass tip over and complete an entire flip. The time taken for an entire flip of ...

0

In general a two-axle vehicle driving off of a abrupt edge will tumble back-over-front on the way down. Which way it lands therefore has to do with how fast it tumbles and how long it tumbles for. The reason for the tumbling is because for the time it takes to drive the distance between the axles, the back axle experiences the normal upwards force from the ...

1

No, cars/vans generally will land front-first or on the roof. This is due to the fact that once the front-end of the car starts going over the edge, a torque (due to gravity) induces a rotation on the vehicle. Most highways and bridges around are small enough that the car/van can only rotate a quarter or half turn (which corresponds to the front and roof ...

0

Not in general. Think of it like this: when the car drives off the cliff, at some point it has a portion of it falling and consequently rotating due to torque exerted by gravity while the other end remains at its original height due to normal forces. As a result, the car will tend to have some sort of rotation as it falls. Thus, the height of the cliff ...

5

It could be possible if the parachute was very large, rigid, shaped like a floating object, and you started descending from the vacuum of space. In this case the parachute would float on top of the atmosphere. It's easier to visualize if you imagine the parachute being a boat and you fell into some water; the boat would float on top of the water and reduce ...

3

Without the ability to change the shape of the parachute, no. With the ability, yes - briefly. A modern square parachute acts as a wing, producing enough lift to slow the descent of the vehicle, but it relies on forward momentum to do so and to remain inflated. If the trailing edge of the parachute is pulled down quickly, the air moving under the wing ...

0

Yes it is possible. The trick is to have a parachute which is large enough that it's Schwarzschild radius extends down to the object it is lifting. Under such a circumstance, the parachute would stop ALL motion of the object it is lifting. PS I just watched Interstellar :D

4

The closest you are going to get is a parachute large enough to slow your descent to the point where you can find lift in rising air and climb away. They exist and are called paragliders! Strictly speaking they are still falling at 1 to 2 metres per second but rely on rising air ( thermals, ridge etc ) to 'fall' slower than a parcel of surrounding air. ...

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..

2

I will answer "yes" if you think out of the box for a parachute, which is a way for a person ejected from a plane to fall on the earth safely. Theoretically, one might design a parachute with a layer of helium so as to match the parachute and person downward gravitational force at a certain height, possibly 4 km above ground so as to avoid mountains, with ...

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 ...

0

For all practical purposes, in the Earth's atmosphere, terminal velocity is less than the speed of sound in water by a large margin ie around 15x. So water will cushion an impact. That Mythbusters episode left something to be desired - accuracy. The world record for a shallow dive into 30cm of water is about 11m in height, meaning an impact velocity of ...

0

No the denser sphere pulls away from the lighter sphere immediately. Just think about the instant they're both dropped and they fall some tiny amount of distance. They both run into the same amount of air molecules in the short distance but the net effect of those air molecules is less on the heavier of the two objects because it has more mass and therefor ...

2

For the record, here's a worked solution: If $r$ is the distance between the two point masses $m_1$ and $m_2$--which start at rest--then as both accelerate towards each other, where $$\frac{d^2r}{dt^2} = -\frac{Gm}{r^2} \ \ \ , \ \ \ \ \ m = m_1 + m_2$$ The first step in solving this equation is the least obvious: Multiply both sides by \$\displaystyle ...

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