What is the Kinetic Energy and Velocity of a Falling Asteroid?

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SUMMARY

The discussion focuses on calculating the kinetic energy and velocity of a 50,000 kg asteroid falling to Earth from a significant distance. The correct approach involves using the gravitational potential energy formula U = -GMm/r, where G is the universal gravitational constant. The kinetic energy imparted to Earth is calculated as K = 3.1 x 10^12 J, and the minimum speed of the asteroid upon impact is determined using the formula v = √(2K/m), yielding a speed of approximately 4,000 m/s.

PREREQUISITES
  • Understanding of gravitational potential energy (U = -GMm/r)
  • Familiarity with the work-energy theorem (W = ΔK)
  • Knowledge of conservation of mechanical energy principles
  • Basic understanding of Newton's law of gravitation
NEXT STEPS
  • Research the universal gravitational constant (G) and its applications in physics
  • Study the derivation and implications of the gravitational potential energy formula
  • Explore conservation of energy in gravitational systems
  • Learn about the implications of large distances on gravitational forces and potential energy
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Students in physics, astrophysics enthusiasts, and anyone interested in gravitational dynamics and energy conservation principles.

clope023
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[SOLVED] Kinetic Energy and Velocity

Homework Statement



The gravitational pull of the Earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the Earth's surface this force is equal to the object's normal weight mg, where g=9.8, and at large distances, the force is zero.

a) If a 50000kg asteroid falls to Earth from a very great distance away, how much kinetic energy will it impart to our planet? You can ignore the effects of the Earth's atmosphere.
Express your answer using two significant figures.

b) What will be its minimum speed as it strikes the Earth's surface?


Homework Equations



Wtot = 1/2m(v2)^2-1/2m(v1)^2

Wgrav = mg


The Attempt at a Solution



a) W = mg = (50000kg)(9.8m/s^2) = 4.9 x 10^5 J (wrong)

b) W = 1/2mv^2, v = \sqrt{2W/m} = 4.4 (definetly wrong)

as you can see I tried using the work energy theorem with the given data, but nothing was right, any help is appreciated.
 
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W = mg is a formula for the weight of something near the Earth's surface (not work).

Hint: Look up the gravitational PE between two bodies as a function of distance.
 
Doc Al said:
W = mg is a formula for the weight of something near the Earth's surface (not work).

Hint: Look up the gravitational PE between two bodies as a function of distance.

but no distance was given, I know Ugrav = mgd, d being distance...
 
clope023 said:
but no distance was given, I know Ugrav = mgd, d being distance...
Ug = mgd only applies near the Earth's surface, which is not the case here. Look for another version of a gravitational PE formula.
 
Doc Al said:
Ug = mgd only applies near the Earth's surface, which is not the case here. Look for another version of a gravitational PE formula.

so I guess you're talking about the gravitational PE formula where the force approaches zero:

U = -GMm/r

but no distance is given so I can't find a radius, I'm assuming you're using:

Ugrav + Wtot = K

but I don't understand how I can get it without a given distance so I can find a radius for the potential energy, do I use the radius of the earth?

edit: so apparently -GMm/r = -1/2mv^2, but again I'm stumped about the radius.
 
Last edited:
tried it with the Earth's radius and got the answer for part a)

GMM/r_earth = K = 3.1x10^12 J w00t!

now for part b)

would it be \sqrt{2K/m}?
 
lol, I'm answering all my own questions.

it is what I just posted above, thanks to Doc Al for pointing me in the right direction with the proper formula.
 
Good work!
 
can anyone summarize how to solve this problem, i do not understand how to use this formula GMm/r

what is G,? 9.8?, does it give you potential energy of the object? doesn't the question ask for kinetic?

thanks for the help
 
  • #10
G is the universal gravitational constant, which appears in Newton's law of gravity. Since mechanical energy is conserved, knowing the change in gravitational potential energy allows you to calculate the change in kinetic energy. Read: http://hyperphysics.phy-astr.gsu.edu/Hbase/gpot.html#ufm"
 
Last edited by a moderator:
  • #11
Use conservation of energy. At very large distances, the asteroid has zero potential energy. We can see this is true because U = k/r (k is a constant), so as r tends to infinity, U tends to zero, so very large r we can neglect U (gravitational energy). Then it becomes trivial to solve using conservation of energy. It starts with at least zero kinetic energy.
 

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