You are explaining why astronauts feel weightless while orbiting in the space shuttle. Your friends respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating the acceleration of gravity 248 km above the Earth's surface in terms of g. (The mass of the Earth is 5.97 1024 kg, and the radius of the Earth is 6380 km.)

The value of g is inversely proportional to the distance from the center of the Earth, squared. The Earth's radius is 6378 km. 248 km above the surface is 6626 km abive.

The value of g at the higher altitude is

g = 9.81*(6378/6626)^2 = 9.09 m/s^2

The astronauts feel weightless because their spacecraft moves around the Earth with the same trajectory, since it is subject to the same acceleration as they are. It is the same reason you would feel weightless in a free-falling elevator. The weightless feeling commences as soon as the rocket thrust is turned off.

To calculate the acceleration of gravity 248 km above the Earth's surface, we can use the equation for gravitational acceleration:

\[ g = \frac{{G \cdot M}}{{r^2}} \]

Where:
- \( g \) is the acceleration of gravity,
- \( G \) is the gravitational constant (approximately \( 6.67 \times 10^{-11} \, \text{{N}} \cdot \text{{m}}^2/\text{{kg}}^2 \)),
- \( M \) is the mass of the Earth (approximately \( 5.97 \times 10^{24} \, \text{{kg}} \)),
- \( r \) is the distance between the object and the center of the Earth.

In this case, we need to calculate the acceleration of gravity at a distance of 248 km above the Earth's surface. To do this, we need to determine the total distance from the center of the Earth to this point. We can calculate this by adding the radius of the Earth (6380 km) to the distance above the surface (248 km):

\[ r = 6380 \, \text{{km}} + 248 \, \text{{km}} = 6628 \, \text{{km}} \]

Substituting the values into the equation, we get:

\[ g = \frac{{(6.67 \times 10^{-11} \, \text{{N}} \cdot \text{{m}}^2/\text{{kg}}^2) \cdot (5.97 \times 10^{24} \, \text{{kg}})}}{{(6628 \, \text{{km}})^2}} \]

Now, let's calculate the value of \( g \):

\[ g = \frac{{(6.67 \times 10^{-11}) \cdot (5.97 \times 10^{24})}}{{(6628)^2}} \]

\[ g \approx 8.63 \, \text{{m/s}}^2 \]

So, the acceleration due to gravity at a distance of 248 km above the Earth's surface is approximately \( 8.63 \, \text{{m/s}}^2 \). This value is only slightly lower than the acceleration due to gravity at the Earth's surface (approximately \( 9.81 \, \text{{m/s}}^2 \)). Hence, it is not the weaker gravity that causes astronauts to feel weightless, but rather the effect of freefall they experience while orbiting the Earth.

To calculate the acceleration of gravity at a certain distance above the Earth's surface, we can use the formula for gravitational acceleration:

g = G * (M / r^2)

Where:
g is the acceleration of gravity,
G is the gravitational constant,
M is the mass of the Earth, and
r is the distance between the center of the Earth and the object.

In this case, we need to calculate the acceleration of gravity 248 km above the Earth's surface. First, we need to convert the radius of the Earth from kilometers to meters:

radius of the Earth (r1) = 6380 km = 6380 * 1000 m = 6.38 * 10^6 m

Then, we need to determine the distance from the center of the Earth to the desired location above its surface:

distance from the Earth's surface (r2) = r1 + 248 km = (6.38 * 10^6 m) + (248 * 1000 m) = 6.38 * 10^6 m + 2.48 * 10^5 m = 6.625 * 10^6 m

Now, we can calculate the acceleration of gravity at this distance above the Earth's surface:

g = (6.67430 * 10^-11 N m^2/kg^2) * (5.97 * 10^24 kg) / (6.625 * 10^6 m)^2

Simplifying the equation:

g = (6.67430 * 5.97) * (10^24 * 10^-11) / (6.625^2 * 10^6^2) N/kg

g = (3.9847291 * 10^13) / (4.378125 * 10^13) N/kg

g ≈ 0.911 N/kg

Now, we can compare this acceleration of gravity to the acceleration due to gravity on the Earth's surface, which is approximately 9.8 m/s^2 or 9.8 N/kg.

Therefore, the acceleration of gravity 248 km above the Earth's surface is about 0.091 times the strength of gravity on the surface (g).

From this calculation, you can conclude that gravity is still present in space, it is not weaker. However, astronauts in the space shuttle experience weightlessness because they are continuously falling towards the Earth while also moving forward in their orbit, resulting in a state of continuous freefall. This freefall creates the sensation of weightlessness, even though the force of gravity is still acting on them.

Can you explain wheightlessness. Do you actually weigh nothing in space or just a percentage of your Earth weight? How much would a 100 lb person weigh on each of the other planets? >>

An astronaut, circling the earth in the Space Shuttle, senses that his apparent weight is zero. He cannot feel the normal earthbound sensation of his feet pressing on the floor below him and the floor holding him upright. Walking in the normal sense is impossible; instead he floats from place to place. This state is often referred to as being in freefall, or in zero gravity. Since gravity in space is never zero, the phenomena that we refer to as zero g is, in reality, the human minds sensitivity to, or the feeling of, apparent weightlessness.
Weightlessnes, as perceived by astronauts in the Space Shuttle, occurs at any altitude where the spacecraft velocity is sufficient to hold the spacecraft in a circular or elliptical orbit. As long as the velocity is sufficient, the gravitational force on the astronaut is exactly balanced by the centrigugal force exerted on his body due to his circular motion about the earth.
The Law of Universal Gravitation states that each particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Expressed mathematically, F = GM(m)/r^2, where F is the force with which either of the particles attracts the other, M and m are the masses of two particles separated by a distance r, and G is the Universal Gravitational Constant. The product of G and, lets say, the mass of the earth, is sometimes referred to as GM or mu (the greek letter pronounced meuw as opposed to meow), the earth's gravitational constant. Thus the force of attraction exerted by the earth on any particle within, on the surface of, or above, is F = 1.40766x10^16 ft^3/sec^2(m)/r^2 where m is the mass of the object being attracted and r is the distance from the center of the earth to the mass. The force of attraction which the earth exerts on our body, that is, the pull of gravity on it, is called the weight of the body, and shows how heavy the body is. Thus, our body, being pulled down by by the earth, exerts a force on the ground equal to our weight. The ground being solid and fixed, exerts an equal and opposite force upward on our body and thus we remain at rest. A simple example of determining this force, or our weight, is to calculate the attractive force on the body of a 200 pound man standing on the surface of the earth. Now the man's mass is his weight divided by the acceleration due to gravity = 200/32.2 = 6.21118 lb.sec^2/ft. The radius of the surface from the center of the earth is 3963 miles x 5280 ft/mile = 20924640 feet. Thus the attractive force on his body is 1.40766x10^16(6.21118)/20924640^2 = 200 pounds. What do you know? The mans weight.
Now lets look at an astronaut inside the Space Shuttle for instance. His so called weightlessness, used to be referred to as being in a state of constant free fall. But since the term "fall" also suggests "down," while the space vehicle is moving in any direction but down, the term "zero g" was ultimately substituted for "free fall." But this clarification often leads to another misconception. The astronaut is not at all experiencing zero g as the earth's gravity is still pulling him down toward the earth. The reason he does not feel this "pull" of gravity is the fact that his velocity through the vacuum of space is creating an equal and opposite centrigugal force on his body exactly cancelling the pull of gravity on his body thus placing him in what is typically referred to as a state of weightlessness. To illustrate, lets look at our 200 pound astronaut as he is hurtling through space in a 250 mile high orbit. His velocity at this altitude is 25,155 feet per second, coincidentally the same as the Space Shuttle :-).
His radius from the center of the earth is 2,224,464 feet. The pull of gravity on his body at this altitude is give by F = 1.40766x10^16(6.21118)/(22244640^2) = 176.7 pounds downward. The outward centrifugal force exerted on his body is given by F = m(V^2)/r = 6.21118(25155^2)/22244640 = 176.7 pounds. Thus he feels no feeling of weight as he would feel on the surface of the earth.
Since attracting masses are present everywhere in the universe; the Sun, the fixed stars, the planets, satellites of the planets, etc., an object will everywhere and at all times be in a gravitational field. Rest and uniform motion are ficticious states and cannot exist anywhere in the universe. To avoid any misunderstanding, it must be pointed out that the gravitational field of any body is infinite. It is however true that, at great distances, the attraction becomes so small that is becomes negligible for all intensive purposes, but it cannot be said that the gravitational field of, any body, has a limit anywhere.
To summarize, gravity in space is never zero and the phenomena that we refer to as zero g is, in reality, the human minds sensitivity to, or the feeling of, apparent weightlessness. An astronaut in the space shuttle, circling the earth, senses that his apparent weight is zero.

The simplest way is to take the value of gravity on the surface of the other planet from any reference source and multiply your weight of 100 lb. by g(P)/g(E) where g(P) is the value of gravity on the planet of interest and g(E) is the value on Earth.
The value of g may also be calculated from the equation of g = GM/r^2 where GM is the Gravitational Constant of the reference body and r is the mean, or equatorial, radius of the reference body. For instance, GM for Pluto is ~1.56259x10^15ft.^3/sec.^2 and r is ~1553.5 miles. So g(P) = 1.56259x10^15/[1553.5(5280)]^2 = ~23.22 ft./sec.^2. This makes your weight of 100 lb. 100(23.22)32.15 = ~72 1/4 pounds.