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.
Step-by-step 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!!!