Understanding Gravity at Various Heights Above the Earth’s Surface

Gravity plays a crucial role in our understanding of physics and has significant implications for our daily lives. One interesting question revolves around the height above the Earth's surface at which the gravitational force is reduced to 81% of its value at the Earth's surface. This article explores the concept and provides a detailed explanation of how to calculate this height.

Calculating the Height Where Gravity is Reduced to 81% of Its Surface Value

The force of gravity at a height (h) from the Earth's surface can be calculated using the formula:

[ g' g left(1 - frac{2h}{R}right) ]

Where:

(g') is the force of gravity at height (h) above the Earth's surface. (g) is the force of gravity on the surface of the Earth (approximately (9.8) m/s2). (h) is the height at which the gravity is to be measured. (R) is the radius of the Earth (approximately (6400) km).

We know that the gravitational force of Earth at sea level is (9.8) m/s2. To find the height where this force is reduced to (81%) of its value, we can use the following steps:

Calculate the reduced force of gravity: (7.9) m/s2) Rearrange the formula to solve for (h): [ g' g left(1 - frac{2h}{R}right) implies frac{g'}{g} 1 - frac{2h}{R} implies 0.81 1 - frac{2h}{6400} ] Solve for (h): [ 0.81 1 - frac{2h}{6400} implies frac{2h}{6400} 0.19 implies 2h 6400 times 0.19 implies 2h 1216 implies h 608 text{ km} ]

Therefore, the height at which the gravitational force is reduced to (81%) of its value at the Earth's surface is approximately (608) km.

Gravitational Pull and Radius Proportions

The gravitational pull decreases with the square of the distance from the center of the Earth. This relationship can be expressed as:

[ frac{1}{r^2} : 1 1 : frac{1}{R^2} ]

If we consider another sphere with radius (R) where (R > r), the gravitational pull on the surface of the Earth can be compared to the pull on a surface of the larger sphere. Using the formula, we find that:

Multiply both sides by (r^2) to get: (R^2 frac{r^2}{3/4}) Solving for (R) gives: (R sqrt{4/3} times 4000) miles ≈ 4600 miles

So, at an altitude of approximately 600 miles, the gravitational pull is (3/4) of what it is on the surface of the Earth.

Gravitational Acceleration at Distance Formula

The gravitational acceleration at the surface of the Earth can be expressed as:

[ g frac{GM}{r^2} ]

Where (G) is the gravitational constant, and (M) is the mass of the Earth. As the distance from the Earth's surface increases, the gravitational acceleration decreases. At some height (R), the acceleration is (0.16g).

Using the formula:

[ frac{GM}{R^2} 0.16 times frac{GM}{r^2} ]

Cancelling out (GM), we get:

[ frac{1}{R^2} 0.16 times frac{1}{r^2} implies R^2 6.25 r^2 implies R 2.5r ]

Substituting the value of (r 6400) km:

[ R 2.5 times 6400 16000 text{ km} ]

Therefore, at an altitude of (16000 - 6400 9600) km, the gravitational force would be (0.16g).

This exploration shows the fascinating relationship between the height above the Earth's surface and the gravitational pull, revealing the beauty and complexity of gravitational forces in our universe.