What would the muzzle-velocity have to be for a two-pound "bullet" to escape Earth's gravity?
146292
11.2 km/s with the following caveats:\n\n"Due to the atmosphere it is not useful and hardly possible to give an object near the surface of the Earth a speed of 11.2 km/s, as these speeds are too far in the hypersonic regime for most practical propulsion systems and would cause most objects to burn up due to atmospheric friction. For an actual escape orbit a spacecraft is first placed in low Earth orbit and then accelerated to the escape velocity at that altitude, which is a little less, ca. 10.9 km/s. The required extra velocity, however, is less because the spacecraft has already been accelerated to about 8 km/s."
1102604
This is an oddly specific question just to be asked out of curiosity.\n\nThe mass of the cannon ball is 13.6 kg, thus it has a weight of about 133.4 Newtons (assume g = 9.81 m/s^2).\nThe force of weight is pointed down...however, since the cannon ball in not in a vacuum and/or not a point mass, it has a non-zero buoyant force acting upward on it.\n\nThe buoyancy force acting on the ball is dependent on the weight of the fluid (water in this case) it displaces.\nIron has a density of about 7874 kg/m^3, thus a 13.6 kg sphere would have a volume of about 1.73 E-3 m^3.\nIf we assume the salt water of the ocean has a uniform density of about 1028 kg/m^3, then the weight of the displaced water exerts a force of about 17.42 Newton’s upward.\n\nThus, the net force acting on (stationary) cannon ball is about 116 Newtons downward.\nAccording to Newton’s 2nd law,\nF = ma\nThis force will cause the 13.6 kg cannon ball to accelerate at a rate of 8.53 m/s^2.\n\nAssuming there is no "water" resistance / drag, then using the formula,\nd = 1/2 at^2,\nwhere the ball stars from rest and experiences a uniform acceleration for a time t as the ball covers d distance,\nThe cannon ball should take about 50.57 seconds to reach the bottom.\n\nHOWEVER,\nJust as there would be air friction if this ball was launched into the air, there is most defiantly "water" friction as this ball accelerated downward....eventually reaching a terminal velocity.\nThe water will exert a drag force upward on the ball, causing its acceleration to be non-uniform.\n\nThe drag force can be calculated as,\nF_d = C_d * ((p * v^2) / 2) * A\nwhere F_d is the drag force, C_d is the drag coefficient, p is the density of the medium which the ball is falling in, v is the instantaneous velocity of the ball, and A is the cross sectional surface area of the ball in the direction of travel.\nIn order to calculate the drag force, one must know the drag coefficient of the cannon ball in water, something which usually must be calculated experimentally.\nThe ball will accelerate non-uniformly until it reaches it terminal velocity in the water, at which time, acceleration will cease and the ball will continue falling at its terminal velocity.\n\nUnless more information is know (the drag coefficient), the best (guess)/estimate of the how long the ball will take is sometime longer than 50.57 seconds...probably on the order of well over a minute since the ball will not be accelerating nearly as long as it did in our mathematical model calculate earlier.
169523
I normally refrain from answering questions that have already been answered by so many people, but in this case, so many of the answers are incorrect, and it might be useful to have a single answer that clarifies what's correct and what isn't.\n\nAn object in freefall in air is subject to two forces. Gravity, which directly proportional to the mass of the object, pulls the object down. Drag acts opposite to the object's movement, and is related to the objects size, its shape, the square of its velocity, and the density of the fluid through which it's traveling. In this case, the drag force acts upwards\n\n^  drag\n|\nO  object\n|\nv  gravity\n\nThe force due to gravity is essentially constant (it increases slightly as the object gets closer to the earth, but this increase is negligible). The force due to drag is related to the size ans shape, which is constant for rigid bodies, and to the speed, which is increasing initally.\n\nThe objects will accelerate until the forces are equal. This occurs at terminal velocity. I would estimate this to be approximately 100 mph. It would take more than 300 ft to reach this speed.\n\nIn a vacuum, there is no drag force, so the acceleration due to gravity (about 9.8m/s^2 near the earth's surface) would continue to accelerate the objects until another force acted on them (e.g. hitting the ground).\n\nSimilarly sized, shaped, and weighted objects such as an egg and an orange could be expected to fall at almost the same rate.\n\nOne final note has to do with a term I used earlier. Neither an orange, nor and egg is really a "rigid object". The egg especially might experience some degree of sloshing inside which could affect its motion. This is why tanker trucks have compartments. I'm not sure if you'd notice this effect, but if you hard boil one egg, you could perform an experiment yourself.\n\nThe bottom line is that you really shouldn't notice any difference in the way the orange or the egg falls.
786994
velocity-height and horizontal movement. 11.2 km/s in order to escape earth's atmosphere (air provides resistance against movement) above 50 km and movement in horizontal direction so that the object falls perpetually towards the earths gravity centre.
314013
"M x V to the power of 2 ÷ 450400 = foot pounds energy"\n"Example: The muzzle energy of a 300 Remington Ultra Mag 180gr Core-Lokt Ultra bullet propelled at 3250 feet per second"\n\n"Step 1: Multiply M (M = bullet weight in grains) times V2 (V2 = the square of bullet velocity in feet per second): 180 x 3250 x 3250 = 1,901,250,000"\n\n"Step 2: Divide the product of step 1 by 450400: 1,901,250,000 ÷ 450400 = 4221 foot pounds of energy."
735664
we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0\n\nwhat we need is the value of acceleration ( remember F=ma )\n\nso the formula we need is\n\nvfina^2=viinitial^2 + 2ad\n304^2=0 + 2*0.86*a\na=(304^2)/(2*0.86)\na=53730.23256\n\nso F=m*a=53730.23256*.0043=231.04 N\n\nForce is 231.04 N\n\n*****************************************************\n\nHere's an alterative if you want to double check  \n\nthe kinetic energy (KE)=1/2 *m*(velocity)^2\n\nthe KE of the bullet must equal the Work needed to fire it\n\nKE=0.5*0.0043*304^2 =198.6944 j\n\nWork=F*d=198.6944\n\nso F=(198.6944)/0.86=231.04 N\n\nAgain we get  force = 231.04 N\n\nSo we're pretty sure this is the correct answer
851042
Weight of the bullet..IE. grain  how far it'll travel @ what velocity
61784
That's an interesting qts. I assume that you wish to invlove pure physics in this qts. Forget all the heat at the core of the earth and also air resistance.\nConsider pure gravitation theory.\n\nIf say the fellow who fell in the hole has mass m\n\nThen force of gravity exerted by the earth is :\n\n  F = GMm/(r*r) ..... Where G - universal gravitation constant\n\n                            M - Mass of the earth\n                            r - distance between earth and body of mass m\n\n          acceleration due to gravity\n            = (1-h/R)g   .... where h is distance of body beneath the earth's surface and g is acceleration due to gravity at the surface of the earth.\n\nSo as the body comes closer to the core of the earth, h tends to R.\nSo the acceleration of the body goes on decreasing very much. So, body may not actually hover, since his velocity will be constant the instant it is at the core.
647242
A rocket doesn't have to fly a certain speed to leave the earth.  A rocket has an engine which means that it needs no minimum speed to get out of the earth's gravitational field.  Escape velocity is based on an object which will not be experiencing powered flight.  A rock thrown upward from the ground will need an initial velocity of 11.2km/s upwards to leave the gravitational field of earth.  A rocket needs to go just fast enough to continue moving (assuming a VERY large amount of fuel to continue engine burn for this duration).
164638
I must bebunk the answerers who think the bullet will be travelling faster without the absence of gravity - the answer is yes and no - yes if compared to a bullet fired vertically on earth or at any angle, no if compared to bullet fired horizontal to the ground on earth.\n\nIf bullet is shot horizontally here on earth, and then shot in space, both bullets will travel at the same speed. The bullet travelling horizontally on earth will have gravity only affect its vertical movement, hence it will be pulled faster and faster towards the ground, but its horizontal velocity will stay the same.\n\nStraightdope has covered this in an article:
1111942
Not necessarily. If your theoretical gun shoots a bullet that reaches escape velocity [1] (the speed something needs to travel to beat Earth's gravity), it can overcome the Earth's gravity [2] and fly off into space.\n\nHowever, I know of no such gun. Only rocket engines [3] are able to do this, because they sustain and increase their speed by burning fuel. A bullet has nothing to keep its speed up once it leaves the gun.\n\nSo the answer is, "Yes, the bullet comes back down." The bullet is always beholden to gravity, even before it's fired. You can view gravity as an invisible string always pointing to the center of the earth, always being pulled. The instant the bullet leaves the barrel of the gun, it will be constantly slowing down due to gravity (and air resistance).\n\nAnother interesting idea is this: If you were to take a gun and fire it horizontally, and at the same instant drop a rock from the same height as the gun, which would hit the ground first? The bullet or the rock? \n\nThe answer is that they both would hit the ground at the same time. The bullet would just be several hundred feet away when it hits. Both are affected by gravity's pull, and approach the ground at the same speed.
604528
Gravity won't let you. You have to go into orbit first, and 1 mph will not do that. You have reach escape velocity to escape earth's gravity, which is 11.2 km/s.
905951
First of all I assume the height of \nbasket above ground is 2+3.05=5.05\nLet v be the required velocity.\nvertical component=vsin40\n=v*.766\nfrom conservation of energy\nheight gained\n=[v*.766]^2/2g=3.05\nv^2=3.05*2*9.81/.766^2\n=101.986\nv=10.09m/sec
145934
You need to go from 0 to 11.2 km/s (earths escape velocity) and you need to overcome earths gravitational pull, which is 9.82 N/KG, in order to do so. You are constantly being accelerated downward with one G (9.92 m/s^2). If you would create a force upwards of one G you would hover (0 G). Let´s say you had a rocket engine that could thrust your spaceship upwards with 2 Gs. How long would it take to reach escape velocity? You use the equation t=v/a where v is 11200m/s and a is 9.81m/s^2 and get 1141.7s. You need to accelerate for 19 minutes to reach escape velocity. Would be sweet if you could do that but it is a long time to keep an engine running. Let´s accelerate twice as fast. Put a as 20m/s^2 and you´ll get 560 seconds or 3.3 minutes. \nOK. So now we´re cooking. How much thrust do you need to get this kind of acceleration? It´s another simple equation: F=ma. We have a (20m/s^2). But you also need m that is the mass of the spacecraft. This we can only guess but lets say a tonne (1000 kg). Then it is pushing downwards onto the launch pad with the force of 1 G wich is 9810 N (9.81x1000). We need to overcome this by adding another G to our thrust so a is 3Gs or 29.81m/s^2 (20 + 9.81). Thus we get 29810 N of thrust for 560 seconds to kick that 1 tonne payload into outer space. That´s 40.5 horsepowers to you non-metric folks (go metric already. A pound isn´t alwasy a pound is it? So measuring thrust in pounds isn´t valid). "But wait. Doesn´t that sound a little too sweet? 40.5 hps? My old Dodge can do better than that". That is correct. You also need fuel to make thrust. The fuel adds weight to the spacecraft which means we need more thrust. Whch means the need for more fuel. Also the craft is exposed to alot of drag from the air. More fuel. More weight. The farther you get from earth the less the value of a G gets but the spacecraft uses up fuel which makes up for the amount of force needed for acceleration. And so on. So this little thought experiment can only take us so far but I´ll stop here anyway. Now you got an idea atleast.
1078996
When bullets are fired straight into the air they go well.....up.  Just like you and me when we jump on a trampoline eventually gravity will get the better of it and it will start coming back down.  Now if that were the only thing to take into account it would just come straight back down and thump you on the head, but because of extra forces caused by air flow and rotation the bullet won't go exactly straight up and down (thankfully).  Now a bullet that's shot straight up is going an average 900 m/s, which means it'll get up about 4,000 meters in the air and when taking into account air resistance it'll come back down with a speed of about 100 m/s.  This is a fraction of the speed but is still fast enough to pierce human flesh.  To answer the final question, the apogee is the highest point that the bullet reaches.
6491
11.3 Km/sec at earth's surface (7 miles/sec)\n\nNOTE that escape velocity is the speed with which you would throw an object so that it does not fall back to earth. Rockets for example have constant propulsion and therefore do not need to travel with escape velocity to go into space.
402349
Normally you can't fire a bullet in space. Theres no air to burn with the powder.\n\nAssuming you do though, theres nothing to slow the bullet down, it would travel the same speed as it did on earth, and actually further as there is no friction or gravity to slow it down.