Altitude of a Rocket Launched into Space

AI Thread Summary
A 4.60 kg rocket is launched upward from Earth at a speed of 9.00 km/s. The relevant equation used to determine the altitude is v = √(2GM/r), where G is the gravitational constant. The calculation yielded a radius of 6.8 x 10^-14 m, indicating the height needed to escape Earth's gravitational pull. The mass of the rocket is irrelevant in this context, as all objects in freefall accelerate uniformly. This discussion highlights the application of escape velocity in determining the altitude necessary for a rocket to leave Earth's gravitational influence.
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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