Could a rock be catapaulted from the ground into outer space?

Question

Without the use of rockets, engines, etc.

Just a very strong catapault, or another VERY strong spring loaded contraption.

Is this possible, or would the rock burn up in the atmosphere?

Thanks everyone. This is a reposted question - thanks for the input - this should clarify.

Answer 1

the rock will burn u due to high friction an gravity pulling it back.Also it has nothing to support it incase the initial energy gets expended.(race in a fast car an after going get down an let the car continue to go on it's own,u know it will come to a halt faster than a car which still has more energy pushing it forward.

Answer 2

Theoretically yes. Burning up is a problem of reentry,not leaving our atmosphere.

Answer 3

if the rock was moving at roughly 7.1 km. per sec.

yes

The phenomenon of escape velocity is a consequence of conservation of energy. For an object with a given total energy, which is moving subject to conservative forces (such as a static gravity fields) it is only possible for the object to reach combinations of places and speeds which have that total energy; and places which have a higher potential energy than this cannot be reached at all.

For a given gravitational potential energy at a given position, the escape velocity is the minimum speed an object without propulsion needs to have sufficient energy to be able to "escape" from the gravity, i.e. so that gravity will never manage to pull it back. For the sake of simplicity, unless stated otherwise, we will assume that the scenario we are dealing with is that an object is attempting to escape from a uniform spherical planet by moving straight up (along a radial line away from the center of the planet), and that the only significant force acting on the moving object is the planet's gravity.

Escape velocity is actually a speed (not a velocity) because it does not specify a direction: no matter what the direction of travel is, the object can escape the gravitational field. The simplest way of deriving the formula for escape velocity is to use conservation of energy. Imagine that a spaceship of mass m is at a distance r from the center of mass of the planet, whose mass is M. Its initial speed is equal to its escape velocity, ve. At its final state, it will be an infinite distance away from the planet, and its speed will be negligibly small and assumed to be 0. Kinetic energy K and gravitational potential energy Ug are the only types of energy that we will deal with, so by the conservation of energy,

    (K + U_g)_i = (K + U_g)_f. \,

Kf = 0 because final velocity is zero, and Ugf = 0 because its final distance is infinity, so

    \frac{1}{2}mv_e^2 + \frac{-GMm}{r} = 0 + 0

    v_e = \sqrt{\frac{2GM}{r}}

Defined a little more formally, "escape velocity" is the initial speed required to go from an initial point in a gravitational potential field to infinity with a residual velocity of zero, with all speeds and velocities measured with respect to the field. Additionally, the escape velocity at a point in space is equal to the speed that an object would have if it started at rest from an infinite distance and was pulled by gravity to that point. In common usage, the initial point is on the surface of a planet or moon. On the surface of the Earth, the escape velocity is about 11.2 kilometers per second (~6.96 mi/s), which is approximately 34 times the speed of sound (mach 34) and at least 10 times the speed of a rifle bullet. However, at 9,000 km altitude in "space", it is slightly less than 7.1 km/s.

The escape velocity relative to the surface of a rotating body depends on direction in which the escaping body travels. For example, as the Earth's rotational velocity is 465 m/s at the equator, a rocket launched tangentially from the Earth's equator to the east requires an initial velocity of about 10.735 km/s relative to Earth to escape whereas a rocket launched tangentially from the Earth's equator to the west requires an initial velocity of about 11.665 km/s relative to Earth. The surface velocity decreases with the cosine of the geographic latitude, so space launch facilities are often located as close to the equator as feasible, e.g. the American Cape Canaveral (latitude 28°28' N) and the French Guiana Space Centre (latitude 5°14' N).

Escape velocity is independent of the mass of the escaping object. It does not matter if the mass is 1 kg or 1000 kg, escape velocity from the same point in the same gravitational field is always the same. What differs is the amount of energy needed to accelerate the mass to achieve escape velocity: the energy needed for an object of mass m to escape the Earth's gravitational field is GMm / r, a function of the object's mass (where r is the radius of the Earth, G is the gravitational constant, and M is the mass of the Earth). More massive objects require more energy to reach escape velocity. All of this, of course, assumes we are neglecting air resistance.

[edit] Misconceptions

Planetary or lunar escape velocity is sometimes misunderstood to be the speed a powered vehicle (such as a rocket) must reach to leave orbit; however, this is not the case, as the quoted number is typically the surface escape velocity, and vehicles never achieve that speed direct from the surface. This surface escape velocity is the speed required for an object to leave the planet if the object is simply projected from the surface of the planet and then left without any more kinetic energy input: in practice the vehicle's propulsion system will continue to provide energy after it has left the surface.

In fact a vehicle can leave the Earth's gravity at any speed. At higher altitude, the local escape velocity is lowe

Answer 4

Yes. Escape velocity is 11.2 kilometers per second, (25000 mph) and if you can get something to reach that speed, in any direction, it is gone.

You won't do it with a catapult, but an electromagnetic track and rail could do it.

Read "The Moon is a Harsh Mistress" by Heinlein.

.

Answer 5

before rockets people tried making big cannons that would shoot high up but it didn't really work

you basically need to use rockets

even with a high powered cannon you couldn't shoot something into space, it needs to keep going and create an orbit...there's also escape velocity etc...

i don't think any sort of catapult would work

Answer 6

Yes, with a catapult capable of greater than 37000 ft/sec release velocity.  

The rock would probably not burn up because the time in the denser part of the atmosphere would only be about 3 sec.

A more detailed analysis required to find out how much greater v is needed; the above # does not consider air drag, which would be very high for the first few seconds of flight.

Answer 7

Surface escape velocity is the speed required for an object to leave the planet if the object is simply projected from the surface of the planet and then left without any more kinetic energy input: in practice the vehicle's propulsion system will continue to provide energy after it has left the surface.

A vehicle can leave the Earth's gravity at any speed. At higher altitudes, the local escape velocity is lower. But at the instant the propulsion stops, the vehicle can only escape if its speed is greater than or equal to the local escape velocity at that position. At sufficiently high altitude this speed can approach 0.

Problems with the catapult method:

- Initial velocity would most likely be insufficient

- Air resistance would slow the projectile considerably

- Gravity is too strong on most points of Earth (higher is better)

- How would you get back?

Answer 8

I would think that gravity woud pull it down.