Calculate the Density of the planet

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SUMMARY

The discussion focuses on calculating the density of a newly discovered planet with a radius of 3000 km, using the time it takes for a stone to fall from a height of 92.5 m. The astronaut measures the fall time as 5 seconds, leading to an acceleration of 2.72 m/s². The calculations involve using gravitational equations, specifically the energy conservation principle and the gravitational force equation, to isolate the planet's mass. The correct approach is to use the planet's radius alone for calculations, as including the rocket height does not significantly affect the results.

PREREQUISITES
  • Understanding of gravitational force equations, specifically DeltaV = -GMm/Rf - (-GMm/Ri)
  • Familiarity with kinematics, particularly the equation .5at² = dY
  • Knowledge of energy conservation principles in physics
  • Basic algebra for isolating variables in equations
NEXT STEPS
  • Study gravitational force calculations using Newton's law of universal gravitation
  • Learn about energy conservation in gravitational fields
  • Explore kinematic equations and their applications in free fall scenarios
  • Investigate how to calculate density from mass and volume in planetary science
USEFUL FOR

Astronomy students, physicists, and anyone interested in planetary density calculations and gravitational physics will benefit from this discussion.

enkerecz
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Homework Statement


An astronaut lands on a new planet of radius R=3000km, and wants to measure its density. In order to do that, he drops a stone of unknown mass from the top of the space shuttle, which has a height of 92.5m and measures the time it takes it to reach the ground. t= 5s.
What is the density of the planet?

Homework Equations


K=.5mv2
DeltaV= -GMm/Rf - (-GMm/Ri) where Rf and Ri are the radius of the planet and the radius of the planet + rocket height respectively.
K=delta V

The Attempt at a Solution


I attempted to solve for the acceleration via kinematics.
I used .5at2=dY and found
a=2.72 m/s2.
Next, I calculated the final velocity using
vf2=sqrt(2adY)=22.433m/s.
I then proceeded to use my Energy equations in order to isolate big M. Unfortunately, my answer is negative and doesn't make sense.. Ehh, actually, the magnitude of it is within reason, but it being negative is what is getting me
answer= -3.77*10^23.. I don't think units cancel properly with what I did :/. Can anyone give me a pointer? I can do the rest if I can just calculate this mass.
 
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What I did, was after I calculated the a from my kinematics,
I set \frac{G(bigM)m}{r<sup>2</sup>}=aplanetm
The little m's cancel leaving me with the ability to calculate M.

The problem now, is which r do I choose? I can choose R of the planet plus rocket height, or R of the planet. It shouldn't make a difference though because the planet's radius is
3*10^6m and the rocket is only 95m tall, correct?
 
That's correct, you can just use R of the planet. And that will give a positive value for M.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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