Many people mistakenly believe that the astronauts who orbit Earth are "above gravity." Earth's mass is 6×1024kg, and its radius is 6.38×106m (6380 km).

Use the inverse-square law to find a height above Earth's surface at that the force of gravity on a shuttle is about 89 % that at Earth's surface.

Well, it's no surprise that people think astronauts are "above gravity." I mean, they're floating around in space, doing somersaults and whatnot. But let's clear up this misconception with a bit of math and humor, shall we?

Using the inverse-square law of gravity, we know that the force of gravity decreases with the square of the distance. So, if we denote the force at the surface as F(surface), and the force at a height h above Earth's surface as F(h), we have the equation:

F(surface) / F(h) = (distance(surface)^2) / (distance(h)^2)

Now, we want the force at a height h to be about 89% of the force at the surface. Let's plug in the numbers and solve for h:

0.89 = (distance(surface)^2) / (distance(h)^2)

Simplifying a bit:

(distance(h)^2) = (distance(surface)^2) / 0.89

Taking the square root of both sides:

distance(h) = square root(distance(surface)^2 / 0.89)

Now, substituting the values for Earth's mass and radius:

distance(h) = square root((6.38×10^6 m)^2 / 0.89)

Calculating that bad boy for you...

*BEEP BOOP BEEP BOOP*

Drum roll, please! According to my calculations, the height above Earth's surface where the force of gravity on a shuttle is about 89% that at Earth's surface is approximately 10,103 kilometers!

So, unless you've got a rocket ship handy, you're stuck down here on Earth, where the gravity is real and the memes about space being "above gravity" are just floating around. Keep your feet on the ground, my friend! Or at least, as close to the ground as you can.

To find the height above Earth's surface at which the force of gravity on a shuttle is about 89% of that at Earth's surface, we can use the inverse-square law of gravity.

The inverse-square law states that the force of gravity between two objects is inversely proportional to the square of the distance between their centers. Mathematically, this can be expressed as:

F = G * (m1 * m2) / r^2

Where:
F is the force of gravity,
G is the gravitational constant (approximately 6.67430 x 10^-11 N.m^2/kg^2),
m1 and m2 are the masses of the two objects, and
r is the distance between their centers.

In this case, we want the force of gravity on the shuttle at a certain height (let's call it h) to be 89% of the force at Earth's surface.

Let's assume that the mass of the shuttle is negligible compared to the mass of the Earth. So we only need to consider the mass of the Earth in the equation.

F_surface = G * (m_shuttle * m_earth) / r_surface^2

F_height = G * (m_shuttle * m_earth) / (r_surface + h)^2

We know that F_height = 0.89 * F_surface.

Therefore, we can set up the following equation:

0.89 * F_surface = G * (m_shuttle * m_earth) / (r_surface + h)^2

To simplify the calculation, we can cancel out the common factors:

0.89 * (m_shuttle * m_earth) / r_surface^2 = (m_shuttle * m_earth) / (r_surface + h)^2

Now we can solve for h:

(0.89 * r_surface)^(1/2) = r_surface + h

Taking the square root on both sides:

(0.89 * r_surface)^(1/2) - r_surface = h

Now we can substitute the known values:

r_surface = 6.38×10^6 m

Calculating:

(0.89 * 6.38×10^6)^(1/2) - 6.38×10^6 = h

h ≈ 1.57×10^6 m

So, the height above Earth's surface at which the force of gravity on a shuttle is about 89% of that at Earth's surface is approximately 1.57 million meters or 1,570 kilometers.

To find the height above Earth's surface at which the force of gravity on a shuttle is about 89% of the force at Earth's surface, we can use the inverse-square law of gravity.

The inverse-square law states that the force of gravity between two objects is inversely proportional to the square of the distance between their centers. Mathematically, it can be expressed as:

F = G * (m1 * m2) / r^2

Where:
F is the force of gravity,
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2),
m1 and m2 are the masses of the two objects, and
r is the distance between their centers.

In this case, the two objects are the shuttle and Earth. We want to find the height (r) above Earth's surface at which the force of gravity is about 89% of the force at Earth's surface.

Let's denote the force of gravity at Earth's surface as F_surface and the force of gravity at the desired height as F_height. We can set up the following equation:

F_height = 0.89 * F_surface

Now, let's plug in the known values for Earth's mass (6×10^24 kg) and radius (6.38×10^6 m) to calculate the force of gravity at Earth's surface.

F_surface = (G * m_shuttle * m_earth) / r_surface^2

Here, we don't know the mass of the shuttle (m_shuttle), so we can cancel out the mass by taking the ratio of the forces:

F_height / F_surface = (G * m_shuttle * m_earth) / (r_height^2) / ((G * m_shuttle * m_earth) / (r_surface^2))

Simplifying the equation, we get:

F_height / F_surface = (r_surface / r_height)^2

Now, we can plug in the given percentage (89%) into the equation:

0.89 = (r_surface / r_height)^2

Taking the square root of both sides, we have:

√0.89 = r_surface / r_height

Rearranging the equation to solve for r_height:

r_height = r_surface / √0.89

Substituting the given values, Earth's radius (r_surface) is 6.38 × 10^6 m:

r_height = 6.38 × 10^6 m / √0.89

Evaluating this expression, we find:

r_height ≈ 8.29 × 10^6 m

Therefore, the height above Earth's surface at which the force of gravity on a shuttle is about 89% of the force at Earth's surface is approximately 8.29 × 10^6 meters or 8,290 kilometers.

(6.38E6)² / .89 = (6.38E6 + h)²