The physics of putting a rocket into space

**Felix83** · Feb15-05, 12:03 AM

If a rocket is launched into space in attemt to break orbit and continue into deep space, how do you determine when the rocket "breaks" the gravitational pull of the earth? Since gravity extends infinatly, how do you know when you are at a point where if you kill the engines, it will continue to travel into space? Gravitational strength decreases exponentially right? So would the point be at a certain distance from earth at a certain velocity so as the limit of the velocity function as displacement approaches infinity is nonzero?

**Andrew Mason** · Feb15-05, 12:22 AM

Originally Posted by Felix83

If a rocket is launched into space in attemt to break orbit and continue into deep space, how do you determine when the rocket "breaks" the gravitational pull of the earth? Since gravity extends infinatly, how do you know when you are at a point where if you kill the engines, it will continue to travel into space? Gravitational strength decreases exponentially right? So would the point be at a certain distance from earth at a certain velocity so as the limit of the velocity function as displacement approaches infinity is nonzero?

That is the idea. Think of it in terms of energy. Does the energy required to go from the surface of the earth to an infinite distance approach a limit? What is that limit? (hint: integrate the force x distance from the surface to infinity).

Once you know the maximum energy required to get to any distance, you just have to give the rocket that much kinetic energy to escape.

AM

**Felix83** · Feb15-05, 11:23 AM

cool, so at what rate does the gravitational field strength decrease? so i would have to incorporate this rate into a function of the force required to overcome gravity and set it up as a function with respect to x (x being distance) and then setup the integral from 0 to infinity, and then take the limit of the integral to solve it? lol thats funny we just learned improper integrals in my calc 2 class the other day

**Andrew Mason** · Feb15-05, 11:29 AM

Originally Posted by Felix83

cool, so at what rate does the gravitational field strength decrease? so i would have to incorporate this rate into a function of the force required to overcome gravity and set it up as a function with respect to x (x being distance) and then setup the integral from 0 to infinity, and then take the limit of the integral to solve it? lol thats funny we just learned improper integrals in my calc 2 class the other day

What is the equation for gravitational force (Newton's Law of Universal Gravitation)?

Use that formula for Force in the integration:

$LaTeX Code: W = \\int_{R}^\\infty F(r) dr$

AM

**Felix83** · Feb15-05, 02:33 PM

newtons universal gravitation formula is F = G*(m1*m2)/r^2

lets say you have a 100kg rocket so m2 = 100kg

the mass of the earth is 5.98 x 10^24 kg so m1 = 5.98 x 1024 kg

G is the gravitatonal constant which is 6.67 x 10^-11 N m2/kg2

r is our variable, but we need a starting point
lets say its at sea level where the radius is 6.37 x 10^6 m

so we are going to have the limit as b -> infinity
of the integral of (3.99*10^16)/r^2) dr from 6.37 x 10^6m to b.

the integral before plugging in the limits of integration is (-3.99*10^16)/r

when we plug in the limits of integration we have

(3.99*10^16) / (6.37*10^6) - (3.99*10^16) / b

if we take the limit as b approaches infinity, the second term tends to zero so the limit equals

(3.99*10^16) / (6.37*10^6)

which is approximatly 6.3 Billion Joules! holy ****! thats alot of work to put a 100kg rocket into space !

so to simplfy that if you wanted to use a different weight and starting point,
you could say that the work needed to launch an object into space is

W = (G * Me * m) / r

where G is the gravitational constant, Me is the mass of the earth, m is the mass of the object, and r is the starting distance from the center of the earth. if you plug in the knowns you can get rid of a couple variables and simplify it farther...

W = ((6.67 x 10^-11) * (5.98 x 10^24) * m) / r

W = 3.99x10^14 * m/r

**Andrew Mason** · Feb15-05, 07:26 PM

Originally Posted by Felix83

If we take the limit as b approaches infinity, the second term tends to zero so the limit equals

(3.99*10^16) / (6.37*10^6)

which is approximatly 6.3 Billion Joules! holy ****! thats alot of work to put a 100kg rocket into space !

It is. But consider that in order to deliver that kinetic energy to the space ship, the rocket has to blast matter (gasses) at high speed toward the earth. Those gases have a lot of energy. And the rocket has to carry that fuel with it while it is blasting away from earth, so much of the rocket energy is used in pushing fuel away from the earth that will later be pushed back toward the earth. 6.3 billion joules is a small fraction of the actual energy needed.

AM

**Felix83** · Feb15-05, 08:27 PM

so basically your saying that the weight of the fuel to carry this 100kg rocket into space would comprise most of the mass making it significantly more than 100kg?

on a side note, just for fun i calculated how fast (neglecting air resistance) any 100kg object would have to start at to get it into space. like if u just took a 100kg sphere and shot it with some initial velocity straight up. In order for it to not come back to earth the initial velocity would have to be around 24000 mph !

**HallsofIvy** · Feb16-05, 10:27 PM

Originally Posted by Felix83

so basically your saying that the weight of the fuel to carry this 100kg rocket into space would comprise most of the mass making it significantly more than 100kg?

on a side note, just for fun i calculated how fast (neglecting air resistance) any 100kg object would have to start at to get it into space. like if u just took a 100kg sphere and shot it with some initial velocity straight up. In order for it to not come back to earth the initial velocity would have to be around 24000 mph !

That's a "classical" problem. Assuming an object (of any mass) is given an initial speed that is then to carry it to an infinite distance, that is the "escape velocity". You calculate the potential energy difference between the surface of the earth and "infinitely distant" (which is finite). The kinetic energy at the surface of the earth must be equal to that.