196k views
1 vote
See question in attached photo.
Answer question 4b and 5c​

See question in attached photo. Answer question 4b and 5c​-example-1

1 Answer

2 votes

9514 1404 393

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

Explanation:

4b

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

__

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

__

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

____

5c

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

__

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%.)

User Radheya
by
7.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories