4. (4 points) Planet Spaceball is a sphere of radius
2 10^8 cm., with no atmosphere. The gravitational force on a 1 gram
object at its surface is 400 dynes (c.g.s units of force). A golf ball on this
planet, weighing 1 gram, is hit at an angle of ¹/3 with respect to the surface
at speed 10^5 cm/sec. What height does it attain?
Solution. Use conservation of energy. The angle introduced an unintended
subtlety, for which reason full credit will be given for more than one
answer, as long as the ideas and calculations were substantially correct.
The easiest simplifying assumption is that the shot is vertical, so that the
energy at the surface of the planet is
(1/2) m v^2 - GMm/rho0
= 5 10^9 - rho0 (GM/rho0^2)
= 5 10^9 - 2 10^8 400)
An alternative for full credit would be to take as the kinetic energy
(1/2) m v^2 (sin(pi/3))^2 = 5 10^9 * (3/4)
The correct physical energy is a bit more complicated than this, because
of the rotational component of the energy.
In[20]:=
Energy = 5 10^9 - 2 10^8 (400)
Out[20]=
-75000000000
It is convenient to write the potential energy this way since we know the
magnitude of the force is
GM/rho^2 = 400. I also used the fact that m=1.
At the top of the orbit, v = 0, so the total energy is
- GM/rhomax = (-GM/rho0) * (rho0/rhomax)
= - 8 10^10 * (2 10^8/rhomax)
In[21]:=
Solve[Energy == - 8 10^10 * (2 10^8/rhomax), rhomax]
Out[21]=
640000000
{{rhomax -> ---------}}
3
Up to Test 3 solutions