What is the Launch Speed of Debris Ejected from Io's Volcanoes?

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Homework Help Overview

The discussion revolves around calculating the launch speed of debris ejected from the volcanoes on Jupiter's moon Io, which is noted for its extreme volcanic activity. Participants explore how the height of the ejected material on Io relates to its potential height on Earth if launched with the same speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation of energy to relate the height and launch speed, questioning the appropriate equations and values for gravitational acceleration on Io. There are attempts to derive the launch speed and check calculations related to maximum height on Earth.

Discussion Status

Several participants have provided guidance on the equations to use and have checked each other's calculations. There is an ongoing exploration of the correct values for gravitational acceleration and the implications for the launch speed. Some participants are verifying their understanding and calculations, leading to productive dialogue.

Contextual Notes

There is a focus on ensuring the correct units are used for gravitational acceleration, as well as clarifying assumptions about the conditions on Io versus Earth. The discussion reflects a mix of understanding and confusion regarding the application of physics principles in this context.

dREAPER
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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500km (or even higher) above the surface. Io has a mass of 8.94×10^22 [kg] and a radius of 1815km . Ignore any variation in gravity over the 500km range of the debris.

How high (in km) would this material go on Earth if it were ejected with the same speed as on Io?

I've calculated g from g = GM/R^2. What equation would I use to calculate the launch speed, v_0, for y= 500,000 m?
 
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Perhaps consider conservation of energy:

mg_{Io}h = \frac{mv^2}{2} \implies v^2 = 2g_{Io}h

Where g_{Io} is the acceleration due to gravity on Io.
 
Coto said:
Perhaps consider conservation of energy:

mg_{Io}h = \frac{mv^2}{2} \implies v^2 = 2g_{Io}h

Where g_{Io} is the acceleration due to gravity on Io.

I don't think we've covered that equation yet. Anything else that you can think of? This chapter covers Newton's Law of Gravitation.
 
You can derive this equation under constant acceleration considerations, but the result is the same. Specifically, the equation is:

2a\Delta y = v_f^2 - v_0^2

In your situation you should see that you know what a is and you know what v_f is.
 
V_f = 0 since it's falling back down?
I plugged my values, and got 55.4223 km. Wasn't correct though. Is my v_f value correct?
 
At the maximum height right before it starts to fall down the velocity should be zero, so yes v_f = 0.

Please write out what you used for your equation replacing a and v_f with what they should be.

What do you find for g_{Io}?

What do you find the initial velocity to be on Io?

What equation do you use to solve for how high it goes on the Earth?
 
g= G_m/R^2 which came out to 1.81*10^6 m/s^2

From there I plugged my variables into 2a\Delta y = v_f^2 - v_0^2

2(-1.81*10^6) 500,000 = 0 - v_f^2 = 1.08*10^6 m/s

Then I replaced a with g of earth, 9.81, since I calculated v_0.

2(-9.81) (delta y) = 0 - (1.08*10^6) = 55.4223 km
 
It looks good dREAPER. Please check over your units (does your answer for g_{Io} make sense?) and please check over your calculations one more time (what is v_0?)

You should end up with ~92km.
 
Coto said:
It looks good dREAPER. Please check over your units (does your answer for g_{Io} make sense?) and please check over your calculations one more time (what is v_0?)

You should end up with ~92km.

Ahh thank you. I missed the units for gravity (it was in km, not m). Calculated it and got 92.3469 m/s^2. Was correct.
 

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