Calculating Escape Velocity for Reaching Altitude of 800km

[naeblis]

Q: In the absenceof air resistance, the least speed with which a body must be projected vertically upward from the Earth's surface if it is to reach an altitude of 800km is...

800km = 8.00x10^5 m
radius of Earth = 6.37x10^6 m
mass of Earth = 5.99x10^24 kg

Kf +Uf = Ki + Ui
0 + 0 = 1/2 mv^2 + (- G Me m / Re + 800km )
1/2 mv^2 = G Me m / Re +800km
m drops out
v = square root of [ (2)(6.67x10^-11)(5.99x10^24)/(7.17x10^6) ]
v = 1.06x10^4 m/s

is this right? thanks in advance for the help

 

[Dale]

Look at your Uf and Ui. Are you sure that Uf is 0? Are you sure about that expression for Ui?

-Dale

 

[Astronuc]

The change in kinetic energy = change in gravitational potential energy if one neglects air resistance (i.e. nonconservative forces).

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

So the objective is to find the initial velocity at which a projectile will leave the Earth's surface, r₁ = r_e, and travel to a radius, r₂ = r_e + 800 km.

So there is an initial KE and GPE and a final KE and GPE, the total energy (KE + GPE) being equal. KE = kinetic energy and GPE being gravitational potential energy.

The minimum speed coincides with KE_f = 0, and thus KE_i = 1/2 mv_i² = \DeltaGPE.

 

[gunblaze]

Your initial potential energy muz include the energy of the body on the surface of the earth... It's not zero.

 

[naeblis]

ok i think i have it now

Kf + Uf = Ki + Ui
0 [because its a minimum velocity] + ( -G Me m / Re +800km ) [potential at 800km] = 1/2 m v^2 + ( -G Me m / Re ) [potential at the begining, the surface of the earth]

then through some algebra we can solve for v

2 [ {-(6.67x10^-11)(5.98x10^24)/(6.37x10^6)+(8.00x10^5)} + {(6.67x10^-11)(5.98x10^24)/(6.37x10^6)} ] = v^2

v = 3.74x10^3 m/s is this right?

hoping my arithmetic is right but i think i got it thanks in advancefor verification!

 

[PhY_InTelLecT]

Yup.. tat shld be the correct ans.