Question

You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating how much weaker gravity is 450 km above the Earth's surface. Express your answer using two significant figures.

Answer 1

Answer:

Only 9% weaker

Explanation:

Because this is where most stuff that people do in space takes place. So, um, here we're at a radius of the earth plus 300 kilometers. You may already be seeing why this isn't going to have much effect if this were except the 6.68 times, 10 to the sixth meters. And so the value of Gout here. You know, Newton's gravitational constant times, the mass of the Earth divided by R squared for the location we're looking at. And so this works out to be 8.924 meters per second squared, which is certainly less than it is at the surface of the earth. However, this is only 9% less than acceleration for gravity at the surface. So the decrease in the gravity gravitational acceleration of nine percent not really going toe produces a sensation of weightlessness.

Answer 2

Answer:

The gravity at the altitude of 450 km above the earth's surface is (0.87 × the normal acceleration due to gravity); that is, 0.87g.

This shows that the gravity at this altitude isn't as weak as initially thought; it is still 87% as strong as the gravity on the surface of the earth.

The gravity is only 13% weaker at an altitude of 450 km above the earth's surface.

Explanation:

The force due to gravity that is usually talked about is basically the force of attraction between the planetary bodies such as the earth and objects on its surface.

The force is given through the Newton's gravitational law which explains that the force of attraction between two bodies is directly proportional to the product of the masses of both bodies and inversely proportional to the square of their distance apart .

Let the force of attraction between a body of mass, m on the surface of the earth and the earth itself, with mass M

F ∝ (Mm/r²)

F = (GMm/r²)

G = Gravitational constant (the constant of proportionality)

r = distance between the body and the earth, and this is equal.to the radius of the earth.

This force is what is now translated to force of gravity or weight of a body.

F = mg

where g = acceleration due to gravity = 9.8 m/s²

F = (GMm/r²) = (mg)

g = (GM/r²) = 9.8 m/s²

So, for a body at a distance that is 450 km above the earth's atmosphere, the distance between that body and the centre of the earth is (r + 450,000) m

Let that be equal to R.

R = (r + 450,000) m

Note that earth's radius is approximately 6400 km

r = 6400 km = 6,400,000 m

R = 6400 + 450 = 6850 km = 6,850,000 m

Normal acceleration due to gravity = 9.8 m/s²

9.8 = (GM/6,400,000²)

GM = 9.8 × 6,400,000²

Acceleration due to gravity at a point 450 km above the earth's surface

a = (GM/R²)

a = (GM/6,850,000²)

Note that GM = 9.8 × 6,400,000²

a = (9.8 × 6,400,000²) ÷ (6,850,000²)

a = 9.8 × 0.873 = 8.56 m/s²

a = 87% of g.

The gravity at this altitude above the earth's surface is (0.87 × the normal acceleration due to gravity)

Hope this Helps!!!