Editing Momentum (section)

=== Objects of variable mass ===
{{See also|Variable-mass system}}
The concept of momentum plays a fundamental role in explaining the behavior of variable-mass objects such as a [[rocket]] ejecting fuel or a [[star]] [[accretion (astrophysics)|accreting]] gas. In analyzing such an object, one treats the object's mass as a function that varies with time: {{math|{{var|m}}({{var|t}})}}. The momentum of the object at time {{mvar|t}} is therefore {{math|{{var|p}}({{var|t}}) {{=}} {{var|m}}({{var|t}}){{var|v}}({{var|t}})}}. One might then try to invoke Newton's second law of motion by saying that the external force {{mvar|F}} on the object is related to its momentum {{math|{{var|p}}({{var|t}})}} by {{math|{{var|F}} {{=}} {{sfrac|d{{var|p}}|d{{var|t}}}}}}, but this is incorrect, as is the related expression found by applying the product rule to {{math|{{sfrac|{{var|d}}({{var|m}}{{var|v}})|d{{var|t}}}}}}:<ref name="kleppner135">{{cite book|last1=Kleppner|last2=Kolenkow|title=An Introduction to Mechanics|pages=135–139}}</ref>

<math display="block"> F = m(t) \frac{\text{d}v}{\text{d}t} + v(t) \frac{\text{d}m}{\text{d}t}. \text{(incorrect)}</math>

This equation does not correctly describe the motion of variable-mass objects. The correct equation is

<math display="block"> F = m(t) \frac{\text{d}v}{\text{d}t} - u \frac{\text{d}m}{\text{d}t},</math>

where {{mvar|u}} is the velocity of the ejected/accreted mass ''as seen in the object's rest frame''.<ref name="kleppner135" /> This is distinct from {{mvar|v}}, which is the velocity of the object itself as seen in an inertial frame.

This equation is derived by keeping track of both the momentum of the object as well as the momentum of the ejected/accreted mass ({{math|d{{var|m}}}}). When considered together, the object and the mass ({{math|d{{var|m}}}}) constitute a closed system in which total momentum is conserved.

<math display="block"> P(t+\text{d}t) = ( m - \text{d}m ) ( v + \text{d}v ) + \text{d}m ( v - u ) = mv+m \text{d}v - u \text{d}m = P(t) +m \text{d}v - u \text{d}m </math>