Finding Altitude: Calculating Earth's Gravitational Field Strength

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SUMMARY

The discussion focuses on calculating the altitude above Earth's surface where gravitational field strength is reduced to four-fifths and one-fifth of its surface value. Using the formula g = GMe/re^2, where G is 6.67 x 10^-11 Nm^2/kg^2 and Me is 5.98 x 10^24 kg, participants are guided to understand the relationship between gravitational strength and distance from the Earth's center. The radius of the Earth is given as re = 6.371 x 10^3 km, which is crucial for solving the altitude problems presented.

PREREQUISITES
  • Understanding of gravitational field strength and its dependence on distance
  • Familiarity with the gravitational constant (G = 6.67 x 10^-11 Nm^2/kg^2)
  • Knowledge of Earth's mass (Me = 5.98 x 10^24 kg)
  • Basic algebra and exponent rules for solving equations
NEXT STEPS
  • Calculate gravitational field strength at various altitudes using g = GMe/re^2
  • Explore the concept of gravitational potential energy and its relation to altitude
  • Investigate the effects of altitude on satellite orbits and gravitational forces
  • Learn about the inverse square law in physics and its applications
USEFUL FOR

Students studying physics, educators teaching gravitational concepts, and anyone interested in understanding the mathematical relationships governing gravitational fields and altitudes.

slu1986
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1. (a) Find the altitude above the Earth's surface where Earth's gravitational field strength would be four-fifths of its value at the surface. Assume re = 6.371*10^3 km.

(b) Find the altitude above the Earth's surface where Earth's gravitational field strength would be one-fifth of its value at the surface.


Homework Equations


g= GMe/re^2
G= 6.67*10^-11 Nm^2/kg^2
Me= 5.98*10^24 kg



3. I am so lost when I try to solve this problem. Could someone please guide me in the right direction to solving this problem. I would greatly appreciate it.
 
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slu1986 said:
1. (a) Find the altitude above the Earth's surface where Earth's gravitational field strength would be four-fifths of its value at the surface. Assume re = 6.371*10^3 km.

Hi slu1986! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)

This is a dimensions question …

(like, if the radius is multiplied by three, how much is the surface area multiplied by?)

so ask yourself, on what power of r does the gravitational field strength depend? :wink:
 

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