Meteor Accelerating towards the Earth

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SUMMARY

The discussion centers on calculating the height of a 12 kg meteor experiencing an acceleration of 7.2 m/s² as it falls towards Earth. Using the gravitational formula \( g = \frac{G \cdot M_{Earth}}{r^2} \), the user correctly derives the distance from the center of the Earth to the meteor as approximately 7.4 x 10⁶ m. Subtracting Earth's radius of 6.36 x 10⁶ m results in a height of 1.06 x 10⁶ m, or 1060 km, above the Earth's surface. A key insight is that the mass of the meteor is irrelevant for this calculation.

PREREQUISITES
  • Understanding of gravitational force and acceleration
  • Familiarity with Newton's law of universal gravitation
  • Knowledge of basic algebra and square root calculations
  • Concept of Earth's mass and radius
NEXT STEPS
  • Study gravitational equations in detail, focusing on \( g = \frac{G \cdot M_{Earth}}{r^2} \)
  • Explore the implications of mass in gravitational calculations
  • Learn about the significance of acceleration due to gravity at different altitudes
  • Investigate the effects of altitude on gravitational force
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Students in physics, educators teaching gravitational concepts, and anyone interested in astrophysics or celestial mechanics.

HarleyM
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Homework Statement



a 12 kg meteor experiences an acceleration of 7.2 m/s^2, when faling towards the earth.

How high above the Earths surface is the meteor?


Homework Equations


mass of earth= 5.98x1024 kg
radius of earth= 6.36x106 m

g=G*MEarth/r2

r=√G*ME/g



The Attempt at a Solution



r=√G*ME/g
=√(6.67x10-11)(5.98x1024)/(7.2)
=7.4x106 m

7.4x106-6.36x106
=1.06 x106 m above the Earths surface, or 1060 km?

Does everything look good? Thanks in advance for any help!
 
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This looks OK.
Did you realize you did not need the mass of the meteor in the calculation??
 

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