Projectile Launch - Earth's Radius

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SUMMARY

The launch speed of a projectile that rises to an altitude equal to three times the Earth's radius is calculated using the formula v_i = √(2GM(1/R_e - 1/(4R_e))). The gravitational constant G is 6.67 x 10^-11 m^3 kg^-1 s^-2, the mass of the Earth M_e is 5.97 x 10^24 kg, and the Earth's radius R_e is 6.37 x 10^6 m. The initial calculation of 5.59 km/s is incorrect because it does not account for the correct potential energy change at the specified altitude. The correct launch speed is approximately 7.91 km/s for an altitude equal to the Earth's radius.

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  • Understanding of gravitational potential energy
  • Familiarity with the gravitational constant (G)
  • Knowledge of Earth's mass (M_e) and radius (R_e)
  • Basic proficiency in algebra and square root calculations
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vworange
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The answer that the book gives for the question:
What is the launch speed of a projectile that rises above the Earth to an altitude equal to the Earth's radius? Ans. 7.91 km/s

Now I have the question:
What is the launch speed of a projectile that rises above the Earth to an altitude equal to three times the Earth's radius?

So I use:
[tex]v_{f} = \sqrt(((2GM_{e})/(4R_{e}))[/tex]

Note:
G = 6.67x10^-11
Me = 5.97x10^34 kg
Re = 6.37x10^6 m

The resulting answer I get is 5.59 km/s. That equation works for an altitude equal to the Earth's radius, but not for my problem. Why not? What am I doing wrong?
 
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Possibly using 4R_{e} instead of 3R_{e}?
 
vworange said:
G = 6.67x10^-11
Me = 5.97x10^34 kg
Re = 6.37x10^6 m

The resulting answer I get is 5.59 km/s. That equation works for an altitude equal to the Earth's radius, but not for my problem. Why not? What am I doing wrong?
The correct energy equation is:

[tex]KE = \Delta PE = GMm(\frac{1}{R_{i}}-\frac{1}{R_{f}})[/tex]

[tex]v_i= \sqrt{2GM(\frac{1}{R_e}-\frac{1}{4R_e})}[/tex]

BTW, the mass of the Earth is 5.97x1024 kg.

AM
 

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