Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
The correct option is c) 83%.
Given :
A shuttle on Earth has a mass of [tex]4.5\times 10^5[/tex] kg.
Shuttle is at height of [tex]6.3\times 10^5[/tex] meters above the surface of the Earth.
Solution :
On the surface of Earth, the weight is
[tex]\rm W = \dfrac{GMm}{R^2}[/tex] ---- (1)
Where,
G is the gravitational constant,
M is the mass of the Earth,
m is the mass of the shuttle,
R is the radius of the Earth.
In orbit the weight is,
[tex]\rm w = \dfrac{GMm}{(R+h)^2}[/tex] ----- (2)
Now we take the ratio of equation (2) to (1),
[tex]\rm \dfrac{w}{W}= \dfrac{R^2}{(R+h)^2}[/tex] ----- (3)
Now put the values of R and h in equation (3) we get,
[tex]\rm \dfrac{w}{W} = 0.83[/tex]
= 83%
Therefore, the correct option is c) 83%
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https://brainly.com/question/2987794?referrer=searchResults
