How can I calculate the escape velocity of a projectile from a planet's surface?

[CollectiveRocker]

I've been thinking about this for a while now, and I was wanting to know what you guys think. The problem states: Show that a projectile with mass "m" can escape from the surface of a planet if it is launched vertically upward with a kinetic energy greater than MgR, where R is the the planet's radius; ignore air resistance. How do I even start? Its in the chapter deal with "thermal properties of matter; thus the ideal-gas law and van de waals equation". Any ideas?

 

[ashfaque]

no...
go to the index of the book and search for "escape velocity"

 

[CollectiveRocker]

I already know that escape velocity is the velocity which it takes to escape a planet's gravitational pull, for example, Earth is around 22,500 mph.

 

[ashfaque]

okke
escape velocity ,v= (2 * G * M / R) ^(1/2) ...(1)

Kinetic energy , KE = 1/2 mv^2

[
now g = GM/R^2
therefor GM = gR^2 ... (2)
]

KE = 1/2 m (2 G M / R) [ from ... (1) ]
= GM * m / R
= gR^2 *m / R [ from 2 ]
= mgR

M maybe is m

 

[CollectiveRocker]

yes M is m, but how does that solve the problem?

 

[CollectiveRocker]

So what do i do now?

 

[CollectiveRocker]

If M = m ,than what do i do?

 

[teclo]

I've been thinking about this for a while now, and I was wanting to know what you guys think. The problem states: Show that a projectile with mass "m" can escape from the surface of a planet if it is launched vertically upward with a kinetic energy greater than MgR, where R is the the planet's radius; ignore air resistance. How do I even start? Its in the chapter deal with "thermal properties of matter; thus the ideal-gas law and van de waals equation". Any ideas?

derive the equation for escape velocity

K(i) = -U(g)

you have a given K(i) is it > -U(g)?

we haven't covered the ideal gas law or van der waals equations yet, though.

 

[CollectiveRocker]

I don't understand what you are saying. Please explain if possible in laymans terms.

 

[Justin Lazear]

This problem has nothing to do with the Ideal Gas Law and such. It's simple classical Newtonian physics.

Use an energy argument. Find the energy of an object that has escaped from the gravitational field of Earth (hint: it's 0), and find the energy of the car on the ground. This two energies must be equal.

Just keep in mind that the car on the ground has two energies associated with it, the potential energy due to the gravitational field (this energy is negative), and the kinetic energy due to the movement of the car.

--J

 

[teclo]

I don't understand what you are saying. Please explain if possible in laymans terms.

well think about escape velocity and how you can derive it from conservation of energy

your initial velocity is the escape velocity, so at a certain time when you're so far away from the planet that your velocity is zero, and its force on you is zero.

in that case you've got

K(i) + U(g) = 0

where K(i) is your initial kinetic energy and U(g) is the potential gravitational energy.

by simple algebra

K(i) = -U(g)

if you calculate this you can derive the formula for escape velocity.

now, you know that you have a kinetic energy greater than MgR (or something like that, right?)

so substitute your values for the equation above

your K(i) > -U(g)

this is saying that your initial kinetic energy is greater than the potential gravitational energy of the planet.

does that make sense? i think it would work