Projectile Motion - Equivalent javelin throw performed on the moon

AI Thread Summary
To determine the equivalent javelin throw on the moon, the release angle of 35 degrees is recommended, and the range can be calculated using the gravitational difference, as the moon's gravity is approximately 1/6 that of Earth's. While some suggest that the range on the moon will be six times that of Earth, this simplification may overlook other factors. A proper approach involves deriving an expression for the projectile's range, assuming the same launch velocity on both celestial bodies. Additionally, while drag and atmospheric resistance are relevant on Earth, they are negligible on the moon, potentially affecting optimal launch angles. Ultimately, the problem encourages a deeper exploration of projectile motion principles rather than relying solely on simplified ratios.
rachel1234
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Homework Statement


Experts recommend a release angle of 35 degrees for javelin throwing, use this as the release angle. Determine what the length of the equivalent throw on the moon would be. Clearly communicate the procedure you followed, and fully justify your answer.


Homework Equations





The Attempt at a Solution


I'm not too sure about the information provided, as it hasn't given a velocity or a range of the projectile. But I guess a general rule for the conversion between a throw on Earth and its equivalent on the moon can still be determined?
I have read some solutions which say that the range will just be six times whatever it was on Earth, as the moon's gravity is approx. 1/6 of Earth's gravity.
I don't know whether it's just me overthinking, but that sounds way too simple!
 
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rachel1234 said:

Homework Statement


Experts recommend a release angle of 35 degrees for javelin throwing, use this as the release angle. Determine what the length of the equivalent throw on the moon would be. Clearly communicate the procedure you followed, and fully justify your answer.


Homework Equations





The Attempt at a Solution


I'm not too sure about the information provided, as it hasn't given a velocity or a range of the projectile. But I guess a general rule for the conversion between a throw on Earth and its equivalent on the moon can still be determined?
I have read some solutions which say that the range will just be six times whatever it was on Earth, as the moon's gravity is approx. 1/6 of Earth's gravity.
I don't know whether it's just me overthinking, but that sounds way too simple!

Hi rachel1234, Welcome to Physics Forums.

You should be able to derive an expression for the range of the projectile which will allow you to prove or disprove your conjecture. Assume that the javelin is launched with the same velocity V in both cases, and that the accelerations due to gravity are ##g_e## and ##g_m##.
 
What gneil said.

Write the equation for the distance on earth. It will contain g. Replace g with 1/6 g.
 
Okay, I think I'm alright with that part now. Thank you!
Also, would I have to discuss the effects of drag, resistance, etc.? The javelin is already a rather streamlined design to reduce the effects of such forces.
 
35 degrees is, presumably, optimal because of the drag in Earth's atmosphere. On the moon, 45 degrees is likely much better, but you're told to use 35. In principle, you could use the optimality of 35 degrees to deduce something about the drag on Earth, and adjust for the lack of drag on the moon. But I suspect that's way beyond the intent of the question.
 
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