# Math for a baseball spaceship

**URL:** <https://boards.straightdope.com/t/math-for-a-baseball-spaceship/967738>\
**Category:** Factual Questions\
**Created:** [July 11, 2022, 1:45am UTC](https://boards.straightdope.com/t/math-for-a-baseball-spaceship/967738 "2022-07-11T01:45:18Z")\
**Posts on this page:** 1\
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**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [July 11, 2022, 7:57am UTC](https://boards.straightdope.com/t/math-for-a-baseball-spaceship/967738/9 "2022-07-11T07:57:07Z")

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Just treat it as continuous. With a million baseballs, it’ll be very close to the same answer.

You haven’t said how many baseballs/second you’re shooting out. I’ll assume 1 baseball/sec.

Ship mass m(t) = (157000 - 0.145t). Force is 145 N (0.145 kg \* 1000 m/s / 1 s). Therefore, a(t) = (157000 - 0.145t) / 145.

To get the total delta V, just use the Tsiolkovsky rocket equation. Fairly easy to derive (I did so [here](https://boards.straightdope.com/t/please-explain-the-physics-of-a-spacecraft-launching-from-the-moon-or-from-mars/919189/67)), but just using an [online calculator](https://www.omnicalculator.com/physics/ideal-rocket-equation) is more convenient. Plug in the numbers and get 2571.3 m/s. Note that this is independent of thrust (i.e., balls/sec).

You could maintain constant acceleration (or any other function) by varying the rate at which you shoot out balls, but it won’t change the delta V.

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