How High Must a Rocket Travel for Its Weight to Be 0.4 Times Its Surface Weight?

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SUMMARY

The discussion focuses on determining the altitude at which a rocket's weight becomes 0.4 times its weight on Earth's surface. Using the inverse-square law of gravitation, the relationship between weight and distance from the Earth's center is established. The radius of the Earth is given as 6.37 x 10^6 m, and the acceleration due to gravity is 9.8 m/s². The calculations reveal that the required height can be derived from the equation 0.4 * weight on surface = G * m * M / x², leading to a solution for x.

PREREQUISITES
  • Understanding of gravitational force and the inverse-square law
  • Basic knowledge of physics, specifically Newton's law of universal gravitation
  • Familiarity with algebraic manipulation and solving equations
  • Concept of weight variation with altitude
NEXT STEPS
  • Study the inverse-square law of gravitation in detail
  • Learn how to apply Newton's law of universal gravitation in various contexts
  • Explore the concept of weightlessness and its relation to altitude
  • Investigate the calculations involved in determining gravitational force at different altitudes
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Physics students, aerospace engineers, and anyone interested in the principles of gravitation and rocket science.

ahsanmhamid
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how high does a rocket have to go above the Earth's surface until its weight is 0.4 times its weight on the Earth's surface ? The radius of the Earth is 6.37 * 10 ^ 6m and the acceleration of gravity is 9.8 m/s^2. Answer in units of km.

what i got so far, i need help proceeding:

Applying the inverse-square law:

Final weight = 0.4 g = (1/2.5) g

The square root of 2.5 = 1.581

im stuck here, thanks :)
 
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Wherever have used or set up the inverse square law?
 
weight on surface = G*m*M/radius^2

0.4*weight on surface = G*m*M/x^2

Solve for x.
 

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