Solving for Planet Islander's Radius from a Projectile Launch

[raisatantuico]

Homework Statement

a projectile of mass m is launched vertically from the surface of Planet Islander at a speed that is one-third the escape speed from the surface. If the projectile reaches a maximum height that is a distance h from the surface of the planet, what is the radius of planet islander?

Homework Equations

Escape Speed = square root of G*mass/radius^2

The Attempt at a Solution

Is this the correct equation:

1/3 (G*mass/r + h) = G*mass/r

so the radius is equal to 9h? or is it 3h? or h/9?

I am confused as what to what equations to equate? and where the height comes in?

Please help!

 

[Filip Larsen]

Your formula for the escape speed and none of the solutions you suggest are correct. You need that, and a consideration of the potential and kinetic energies involved at two interesting points of the projectiles flight in order to solve this problem.

 

[raisatantuico]

is energy conserved in this problem? so we can equate energy at the peak and energy just before it is launched?

 

[Filip Larsen]

Yes, mechanical energy is conserved if you make the assumption that there are no atmosphere.

 

[rwisz]

I would like to clarify that the equation you have provided is not correct. The correct equation to use in this scenario is the conservation of energy equation, which states that the initial kinetic energy of the projectile (1/2mv^2) is equal to the final potential energy at its maximum height (mgh). This can be written as:

1/2mv^2 = mgh

Since the speed of the projectile is one-third of the escape speed, we can substitute this into the equation as:

1/2mv^2 = 1/9(GMm/r)

Where G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet. We can then solve for r by rearranging the equation as:

r = 9h(M/m)

Therefore, the radius of Planet Islander is equal to 9 times the maximum height of the projectile multiplied by the ratio of the planet's mass to the projectile's mass.

I hope this helps clarify the correct approach to solving this problem. Keep in mind that in scientific calculations, it is important to use the correct equations and units to ensure accurate results.