Altitude of a Rocket Launched into Space

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SUMMARY

The discussion centers on calculating the altitude a 4.60 kg rocket reaches when launched upward at a velocity of 9.00 km/s. The relevant equation used is the escape velocity formula, v = √(2GM/r), where G is the gravitational constant (6.67 x 10^-11). The calculation results in an altitude of approximately 6.8 x 10^-14 meters, indicating that the mass of the rocket is irrelevant in this context, as all objects in freefall accelerate uniformly regardless of mass.

PREREQUISITES
  • Understanding of gravitational physics and escape velocity
  • Familiarity with the equation v = √(2GM/r)
  • Knowledge of the gravitational constant, G = 6.67 x 10^-11
  • Basic principles of freefall and acceleration
NEXT STEPS
  • Research the implications of escape velocity in orbital mechanics
  • Explore the concept of gravitational potential energy
  • Learn about the differences between suborbital and orbital trajectories
  • Investigate the effects of mass on acceleration in freefall scenarios
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Students of physics, aerospace engineers, and anyone interested in the principles of rocket launches and gravitational effects on motion.

tesla93
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The Problem:

A 4.60 kg rocket is launched directly upward from Earth at 9.00 km/s

What altitude above the Earth's surface does the rocket reach?


Relevant Equations:

v = √ (2GM/r)

The Attempt:

m = 4.60 kg
v = 9km/s = 9000m/s
G = 6.67 x 10^-11

9000^2 = 2(6.67^-11)(4.60)/r

r = 6.8 x 10-14m

I used that formula because it is for escape speed, which I think applies to this question, as it needs to be a certain height above the Earth to escape the pull and stay in orbit.
 
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"M" is the mass of the Earth, not of the rocket. The mass of the rocket doesn't matter because in freefall, all objects accelerate at the same rate, regardless of mass.
 
Oh alright that makes sense. Thank you!
 

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