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Shoemaker-Levy Comet Fragment 9 impacting Jupiter July 22 1994 |
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Jupiter, showing the impact marks from fragments of Comet Shoemaker-Levy 9 in July 1994 |
Because [gravity] will act to pull meteors, comets, and other space debris toward a planet, the effective cross-section for a planet to capture an object is larger than its geometrical cross-section. What is the size of this effective cross-section in terms of the physical qualities of the planet and the situation? What features of the impacting body is it independent of?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
The Force Diagram of the Meteor approaching the Planet
Force Diagram of Meteor and Planet |
The [single-axis torque] about the center of the planet is zero, because the force of [gravity] acts along the same direction as the radius r. About this point, therefore, [angular momentum] is conserved.
Sketch showing Torque |
Mathematical Representation
The force is supplied by a belt around the smaller wheel of radius r (in a 19th century factory, it would probably be a circular leather belt attached to the water wheels). This means that the direction the force is applied along is always tangential to the circumference of the wheel, and hence Torque = r X F = rF
\begin
[ \vec
= \vec
X \vec
= rF = I_
\alpha ]\end
The Moment of Inertia of combined bodies about the same axis is simply the sum of the individual Moments of Inertia:
\begin
[ I_
= I_
+ I_
]\end
The Moment of Inertia of a solid disc of radius r and mass m about an axis through the center and perpendicular to the plane of the disc is given by:
\begin
[ I = \frac
m r^2 ] \end
So the Moment of Inertia of the complete flywheel is:
\begin
[ I_
= \frac
(m r^2 + M R^2 ) ]\end
The expression for the angular velocity and the angle as a function of time (for constant angular acceleration) is given in the Laws of Change section on the Rotational Motion page:
\begin
[ \omega_
= \omega_
+ \alpha (t_
- t_
) ] \end
and
\begin
[ \theta_
= \theta_
+ \omega_
( t_
- t_
) + \frac
\alpha ( t_
- t_
)^2 ]\end
We assume that at the start, ti = 0 , we have both position and angular velocity equal to zero. The above expressions then simplify to:
\begin
[ \omega_
= \alpha t_
]\end
and
\begin
[ \theta_
= \frac
\alpha {t_{\rm f}}^2 ]\end
where
\begin
[ \alpha = \frac
{I_{\rm total}} = \frac
]\end