Download Physics Challenge Question 1: Solutions

Physics Challenge Question 12: Solutions
Part 1
From class, we know that the total mechanical energy is conserved. That
means that ME initial  ME final . So, we need to identify what the energy is just
when the object is launched off the planet, and when it is infinitely far from
the planet.
1
1
2
2
KEi  mvi  mve (The initial velocity is the escape velocity.)
2
2
1
2
KE f  mv f  0 (If it just escapes the planet, its final speed is 0.)
2
PE i  G 
PE f  G 
M m
R
M m
 0 (When it is infinitely far away, r f   and PE f  0 .)
rf
By energy conservation, we get
KEi  PEi  KE f  PE f
1
M m
2
mve  G 
 0 0.
2
R
Now, we can rewrite to solve for ve:
1
M m
2
mve  G 
2
R
1 2
M
ve  G 
2
R
M
2
ve  2  G 
R
2GM
ve 
R
Part 2
Plugging in to our equation, we get:
2  6.67  10 11 kgms 2  6.0  10 24 kg
3
ve 
2
6,360,000m
 1.12  10 4
m
s
 11.2 kms
Part 3
For this part, we must rewrite our equation. (We want to solve for R.)
2GM
R
2
ve
Now, we can use the numbers given to estimate the radius of the black hole:
3
2  6.67  10 11 kgms 2  2.0  10 30 kg
R
 2,964m  3.0km
3.0 108 ms 2
Even though this is just an estimate (our escape velocity equation isn’t
entirely correct for light beams), it still shows just how extremely compact
black holes are: The mass of our sun (about 700,000 km in radius) has been
squeezed into a radius of just 3 km!