# A golf ball being hit on the moon vs Earth

• Jregan
In summary, the conversation discusses using the range equation to calculate the distance traveled by a golf ball on the moon and on Earth. The equation is appropriate in this case because the projectile lands at the same height from which it was launched. The calculations for the distance traveled on the moon and on Earth are 452m and 75.2m, respectively. The conversation also mentions that one of the Apollo astronauts brought golf balls on the moon and hit them with an adapted sampling tool.
Jregan
Homework Statement
A golf ball is hit on the moon which has a free-fall acceleration 1/6 of its value on earth. The ball is hit at a speed of 28m/s at an angle of 35 degrees. How far did the golf ball travel? Ignoring air resistance, how far would it travel on earth?
Relevant Equations
R=Vi^2sin2theta/g
I am just not sure if I did this properly. My professor hasn't really gone over when to use the range equation but I would assume range would equal the distance traveled therefore can be used for this problem. If not the how would I go about solving this?

I did 1/6*9.8=1.63 for g on the moon

Then used R=(Vi^2sin2theta)/g

so on the moon
R=((28^2)sin2*35)/1.63=452m

and on Earth
R=((28^2)sin2*35)/9.8=75.2m

Jregan said:
Homework Statement: A golf ball is hit on the moon which has a free-fall acceleration 1/6 of its value on earth. The ball is hit at a speed of 28m/s at an angle of 35 degrees. How far did the golf ball travel? Ignoring air resistance, how far would it travel on earth?
Homework Equations: R=Vi^2sin2theta/g

I am just not sure if I did this properly. My professor hasn't really gone over when to use the range equation but I would assume range would equal the distance traveled therefore can be used for this problem. If not the how would I go about solving this?

I did 1/6*9.8=1.63 for g on the moon

Then used R=(Vi^2sin2theta)/g

so on the moon
R=((28^2)sin2*35)/1.63=452m

and on earth
R=((28^2)sin2*35)/9.8=75.2m
Looks good.

berkeman
The use of the range equation is appropriate in this case. You can use it whenever the projectile lands at the same height from which it was launched.

Last edited:
Jregan
Simpler calculation without the atmosphere, no drag, but also no allowance required for 'spin' which, with the dimples, may provide some aerodynamic lift and extend range thus.

IIRC, one of the braw Apollo moon-walkers took a golf-ball or two as part of his 'personal allowance', whacked them 'into the rough' with an adapted sampling tool...

Thank you!

## 1. How far would a golf ball travel if hit on the moon compared to Earth?

The distance a golf ball travels is determined by its initial velocity and the force of gravity. On the moon, the force of gravity is about 1/6th of that on Earth, meaning a golf ball would travel much further. Assuming the same initial velocity, a golf ball hit on the moon would travel approximately 6 times further than one hit on Earth.

## 2. Would a golf ball hit on the moon have the same trajectory as one hit on Earth?

No, the trajectory of a golf ball is affected by air resistance and the force of gravity. On the moon, there is no atmosphere and the force of gravity is weaker, so the golf ball would follow a more parabolic trajectory compared to the more curved trajectory on Earth.

## 3. How does the lack of atmosphere on the moon affect a golf ball's flight?

The lack of atmosphere means there is no air resistance to slow down the golf ball. This allows the golf ball to travel further and faster on the moon compared to Earth.

## 4. How would the golf ball's weight differ on the moon compared to Earth?

The weight of an object is determined by the force of gravity acting on it. On the moon, the force of gravity is weaker, so the golf ball would weigh approximately 1/6th of its weight on Earth. This means it would feel much lighter and be easier to hit.

## 5. Would the distance a golf ball travels on the moon vary depending on the location it is hit?

Yes, the distance a golf ball travels on the moon would vary depending on the location it is hit. This is because the force of gravity is not uniform on the moon, so the distance would be affected by the strength of gravity at different locations.

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