A model rocket won't escape Earth's gravity...but...?

Question

Would a model rocket launched from the Moon escape the Moon's gravity?

Answer 1

Suppose the escape velocity of the Earth is 20,000 MPH.
The Moon's gravity is about 1/8 that of the Earth.
20,000 / 8 = 2,500 MPH
Sure, there is no atmosphere on the Moon, but it still takes energy to accelerate a mass to 2,500 MPH. It would have to be a very good model rocket! I think model rockets reach speeds of about 400-500 MPH. 500 x 8 = 4,000 says yes, it would reach escape velocity. But the actual equations are really complicated.
Here is a nice link:
http://my.execpc.com/%7Eculp/rockets/qref.html

From Rocket Equations:
Boost Phase: Velocity at Burnout

* Rocket thrust = T
* Force of gravity = M*g
* Drag force on rocket = 0.5*rho*Cd*A*v^2 = k*v^2
* Net force on rocket = F = T - M*g - k*v^2
* Newton's Second Law: F = M*a = M*(dv/dt) = T - M*g - kv^2
* Collecting terms: dt = M*dv / (T - M*g - k*v^2) = (M / k)*(dv / [q^2 - v^2])
where I've defined q = sqrt([T - M*g] / k)
* Integrating both sides (finite integral from 0 to v) and rearranging:
t = (M / k)*(1 / [2*q])*ln([q+v] / [q-v])

* Simplifying a bit: 2*k*q*t / M = ln([q+v] / [q-v])
Set x = 2*k*q / M and then
* Solve for v:
v = q*[1 - exp(-x*t)] / [1 + exp(-x*t)]

Boost Phase: Altitude at Burnout

* Newton's Second Law Again
F = M*a = M*(dv/dt) = M*(dv/dy)*(dy/dt) = M*v*(dv/dy) = T - M*g - k*v^2
* Rearranging: dy = M*v*dv / (T - M*g - k*v^2)
* Integrating both sides (finite integral from 0 to v) and rearranging:
y = (M / 2*k)*ln([T - M*g - k*v^2] / [T - M*g])

Coast Phase: Distance Travelled from Velocity v to Zero

* Newton's Second Law Yet Again
F = M*a = M*v*(dv/dy) = - M*g - k*v^2
* Rearranging: dy = M*v*dv / (- M*g - k*v^2)
* Integrating both sides: y = [+M / (2*k)]*ln([M*g + k*v^2] / [M*g])

Coast Phase: Time to Velocity Zero

* Newton's Second Law in the Time form
F = M*a = M*(dv/dt) = - M*g - k*v^2
* Rearranging: dt = M*dv / (- M*g - k*v^2)
* Integrating both sides: t = ([M/k]/sqrt[M*g/k])*arctan(v / sqrt[M*g/k])

Simplifies (some) to: t = sqrt(M / [k*g])*arctan(v / sqrt[M*g/k])

Approximation Using Static Rocket Mass

THE rocket equation assumes a dynamic mass m(t) = m0 - (dm/dt)*t, where dm/dt is a constant. When this expression is substituted into the above Second Law equations, they become intractable and must be solved with numerical methods. I therefore use a static expression for the mass M of the rocket. I will call the velocity found using the dynamic expression "vd", and the velocity found using the static expression "vs".

Define

* Vx = exhaust velocity, speed of propellant leaving rocket
* mr = mass of rocket, when EMPTY
* mp = mass of propellant (total)

Then we have

* THE rocket equation (dynamic mass): vd = Vx * ln([mr+mp] / mr)
* The "static rocket mass" equation: vs = Vx * (mp / mr)
* The static equation equivalent to my method of using average rocket mass is:
vs = Vx * (mp / [mr + 0.5*mp]).

Then a measure of the error induced by my method is E = 1 - vs / vd. Let's suppose the extreme case (for a model rocket) that the propellant is half the total weight of the rocket, or mp = mr = m. Then
E = 1 - vs / vd = 1 - [ (m / {m+0.5*m}) / ln({m+m} / m) ] = 1 - [1 / {ln(2) * 1.5}] = 0.04, or 4% error.

You can verify for yourself that for propellants that are somewhat smaller proportions of the rocket mass, the error is much smaller. The propellant has to exceed 67% of the total rocket mass before a 10% error is induced.

So, plug the numbers for your model rocket into the equations, and guesstimate what would happen if it was launched from the Moon!

;-D Have fun, I hate Math!

Answer 2

No it would not, the answers you're getting are ignorant rubbish by people who have no clue. Model rockets generally don't go more than a few hundred feet up. Multiply that by six for the moon...a few thousand feet up at best. Very expensive model rockets made by dedicated amateurs may go 4000 feet high on Earth, but they still aren't powerful enough to escape the moon's gravity. That trick still requires a velocity of over ONE MILE PER SECOND!!!

Answer 3

Again it is not as an answer but to appreciate the high quality of the r esponses. However,all the five 'yeswallas' have overlooked the contigency so ably pointed out by AC Imagined. Some have quoted the argument of escape velocity both ways. And one answers diplomatically keeps its hand on both the possibilities.The array of the replies,even if of high quality, shows how a lack of attention to all the pros and cons of the matter may vitiate the reply.However, I am happy to find that a serious scientific matter has been dealt with the seriousness it demands.Let this trend continue.

Answer 4

No it would not. You must exceed the escape velocity of an object to escape from its gravity. The escape velocity of the Earth is 11.2 km/sec and the escape velocity of the Moon is 2.4km/sec. I don't know how fast a model rocket goes but I will bet it is well under 1km/sec which would be 3600 mph.

See http://en.wikipedia.org/wiki/Escape_velocity

Answer 5

a model rocket cant escape earths gravity because the boost is not powerfull enough to overcome earth's gravity but if you try to launch it in the moon since its gravity is lesser it might on the condition that its boost is greater than of moon's gravity.

Answer 6

A model rocket does not have enough thrust ot reach escape velocity from Earth but it does from the moon.

Answer 7

No. Gravity extends to infinity.
Even the astronauts from Earth who flew to the moon, were still in Earth's gravity. The moon is still in orbit around the Earth isn't it? Earth's gravity keeps it in orbit.

Answer 8

More than likely!

If a model rocket can go 20 miles up on Earth it can leave lunar orbit.

A model rocket, if made big enough and using real rocket fuel (hydrazine) can go pretty far.

A rocket from a sub is only 30 feet long and it can go hundreds of miles.

Answer 9

Maybe?

The escape velocity of the moon is 2.4 km/sec or about 5400 miles/hour. You'd need one big honking model rocket to go that fast even at the 0.1654g of the moon.

Cool idea though.

Answer 10

Yes, most certainly it would. There's not a lot of gravity on the moon. There is more gravity on Earth than there is on the moon. Also, Earth has an atmosphere... the moon does not.