Question

Question:
A 460kg satellite is launched into a circular orbit and attains an orbital altitude of 850km above Earth's surface. Calculate the:


e) additional energy and speed required for the satellite to escape.

Attempt:
I found Ek to be 1.27*10^10 J
Thus, the total energy is -1.27*10^10 J
The binding energy is 1.27*10^10 J.

Wouldn't the additonal energy it needs be the binding energy?
The escape speed would be calculated by root of:

v(esc) = root of 2GM/r

But the answer at the back says otherwis.e... :roll:

Answer 1

So \[Fc=mv^2/r\]
And \[Fg=Gmm/r^2\]
If it's in orbit, then
\[mv^2/r=Gmm/r^2\]
Rearrange and you get
\[v=\sqrt{Gm/r}\]
6378.1 km is the radius of the earth, so the height would be 850+6378.1=7238.1 km.
The mass of the earth is 5.9742*10^24 kg, and the gravitational constant is 6.673*10^-11 m^3/kgs
\[v= 7,421.5 m/s\]
\[v_{escape}=10,495.6 m/s\]
\[v_{difference}=10,495.6-7,421.5=3,074.08 m/s\]

Answer 2

Now that we have the velocities, I think we can find the energies of the satellite. So since \[KE=1/2mv^2\]
and \[U_{gravity}=-Gmm/r^2\]
\[E=KE-U_{gravity}\]
So after some manipulation \[E=-Gmm/2r\]
So...
\[E=-1.26679*10^{10}J\]
And since the final energy by definition is 0 for escape velocity,
The additional energy required is
\[E=-1.26679*10^{10}J\]