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Answer:
4b: 8.6 m/s²; 1.3×10^5 N; 2.1×10^4 N decrease
5c: -1.6×10^12 J; -1.6×10^12 J
Step-by-step explanation:
<h3>4b</h3>
i) The acceleration due to gravity is inversely proportional to the square of the distance between the objects. The distance to the shuttle is ...
1 + (5×10^5)/(6.4×10^5) = 69/64 . . . times the radius of the earth
Then the acceleration due to gravity at the height of the space shuttle is about ...
(10 m/s²)(64/69)² ≈ 8.6 m/s²
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ii) The weight of the space shuttle at that height is about ...
F = ma = (15000 kg)(8.6 m/s²) ≈ 1.3×10^5 N
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iii) The loss of weight will be ...
ΔF = m(a1 -a0) = (15000 kg)(10 m/s² -8.6 m/s²)
= 1.5×10^4×1.4 N = 2.1×10^4 N
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<h3>5c</h3>
i) The gravitational potential energy is given by ...
U = -GMm/r
where M and m are the mass of the earth and the rocket, respectively.
U = -(6.67×10^-11)(6.0×10^24)(2.5×10^4)/(6.4×10^6) ≈ -1.6×10^12 J
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ii) At a height of 3×10^4 m, the denominator in the above expression changes from 6.4×10^6 to 6.43×10^6. This changes the gravitational potential energy by a factor of 6.4/6.43 to -1.6×10^12 J
(Note: we're carrying only 2 significant figures in the result in accordance with the rules for precision in such calculations. The change is noticeable at the level of the 4th significant figure, less than 1/2%.)
