Calculating the Altitude Above Earth's Surface for a 10% Decrease in Gravity

Understanding the behavior of the force of gravity with respect to altitude is crucial for various scientific and engineering applications. One fundamental question is: at what distance above the Earth's surface does the acceleration due to gravity decrease by 10%? This article delves into the mathematical derivation and practical significance of such calculations.

Introduction to Gravitational Force

The acceleration due to gravity at the Earth's surface is approximately ( g 9.81 text{ m/s}^2 ). To find the height above the Earth's surface where gravity is reduced by 10%, we need to use the formula that describes the variation of gravitational acceleration with distance from the Earth's center:

( g frac{GM}{r^2} )

Here, ( G ) is the gravitational constant, ( M ) is the mass of the Earth, and ( r ) is the distance from the center of the Earth. The radius of the Earth, ( R ), is approximately ( 6.371 times 10^6 text{ m} ).

Derivation of the Height

Assuming we want the acceleration due to gravity to be 90% of its surface value, we set up the equation as follows:

( 0.9g frac{GM}{(R h)^2} )

Given that the acceleration due to gravity at the surface is ( g frac{GM}{R^2} ), we can substitute this into the equation:

( 0.9 times frac{GM}{R^2} frac{GM}{(R h)^2} )

By canceling ( GM ) from both sides, we get:

( 0.9 frac{R^2}{(R h)^2} )

Taking the square root of both sides:

( sqrt{0.9} frac{R}{R h} )

Solving for ( h ):

( frac{h}{R h} 1 - sqrt{0.9} )

Multiplying both sides by ( R h ) and simplifying:

( h R (1 - sqrt{0.9}) )

Substituting ( R 6.371 times 10^6 text{ m} ) and ( sqrt{0.9} approx 0.9487 ):

( h approx 6.371 times 10^6 times (1 - 0.9487) approx 344,000 text{ m} )

Therefore, the distance above the Earth's surface where the acceleration due to gravity is reduced by 10% is approximately 344 km.

Alternative Approach

The force exerted by gravity is proportional to the square of the distance from the Earth's center. The formula ( F propto frac{1}{r^2} ) simplifies calculations. At the Earth's surface, the radius is approximately 6,378 kilometers. To achieve a 10% decrease in the gravitational force, we can use the following approach:

( a g times frac{R}{r^2} )

For a 10% decrease, we solve:

( a 0.1g )

This gives us:

( left( frac{r}{R} right)^2 10 )

Thus:

( r^2 10R^2 )

And solving for ( r ):

( r R sqrt{10} )

Substituting ( R 6378 text{ km} ):

( r approx 6378 times sqrt{10} approx 20200 text{ km} )

The height above the surface would then be:

( h r - R approx 20200 - 6378 approx 13822 text{ km} )

Conclusion

To summarize, the distance above the Earth's surface where the acceleration due to gravity is reduced by 10% is approximately 344 kilometers. This level of detail is essential for fields such as satellite orbit calculations, geophysics, and astrophysics. Understanding these principles allows us to make accurate predictions and adjustments in real-world scenarios.