Question

A projectile is fired from the surface of earth of radius R with velocity KVe (where Ve is the escape velocity from surface of earth and K<1). Neglecting air resistance , what is the maximum height of rise from the center of earth?

Answer 1

Well, since K<1, your velocity is less than the escape velocity and the projectile will be in a bound orbit with the Earth. The total mechanical energy (which is a constant, since friction is being ignored here) of the projectile is:\[E=\frac{1}{2}mv(r)^2-\frac{mMG}{r}\]Where r is the distance of the particle from the center of the Earth at a particular time and v is the projectile's velocity at that same distance r. When the projectile is at it's maximum distance from Earth it's velocity is zero. So,\[E=\frac{-mMG}{r _{\max}}\]Giving,\[r _{\max}=\frac{-mMG}{E}\]The negative sign here makes sense, since for bound orbits E<0, making r{max}>0.We can get a value of E by noting our initial conditions. We fired the projectile from the surface of the Earth (r=R, radius of Earth) at a velocity of kve. So,\[E=\frac{1}{2}mk^2v _{e}^2-\frac{mMG}{R}\]But, \[v _{e}=\sqrt{2 \frac{MG}{R}}\]where M is the Earth's mass. Thus,\[E=\frac{1}{2}mk^2(\frac{2MG}{R})-\frac{mMG}{R}\]giving,\[E=\frac{mMG}{R}(k^2-1)\]Ok, almost done. Now let's substitute this value of E into our equation for r{max} above:\[r _{\max}=\frac{-mMG}{\frac{mMG(k^2-1)}{R}}=\frac{-R}{(k^2-1)}\]Again, the negative sign makes sense, since K<1, the denominator will be negative, giving a positive value for r{max}.

Answer 2

Notice that the limit as k approaches 1 from the left =infinity. This makes sense. If k=1, then the projectile escapes the Earth and reaches r=infinity with zero velocity.

Answer 3

thanx a lot!
you're gr8!

Answer 4

\[R \div \left(1-K ^{2} \right)\]

Answer 5

this was my answer after i derived it!!

Answer 6

Yeah, I guess I could have eliminated the negative sign in my answer...doh!