Redo Exercise 238 from Section 1.4, but instead of a weather satellite with a mass of 1400 kg, calculate how many Joules of work it would take to elevate an almond with a mass of 0.0012 kg to the same altitude. Enter your answer as just a number (no units), rounded to two significant digits (remember, that's not the same as two decimal places!)

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Chapter1: Units, Trigonometry. And Vectors
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Redo Exercise 238 from Section 1.4, but instead of a weather satellite with a mass of 1400 kg, calculate how many Joules of work it would take to
elevate an almond with a mass of 0.0012 kg to the same altitude. Enter your answer as just a number (no units), rounded to two significant digits
(remember, that's not the same as two decimal places!)
Transcribed Image Text:Redo Exercise 238 from Section 1.4, but instead of a weather satellite with a mass of 1400 kg, calculate how many Joules of work it would take to elevate an almond with a mass of 0.0012 kg to the same altitude. Enter your answer as just a number (no units), rounded to two significant digits (remember, that's not the same as two decimal places!)
(238) F(x) = G₁MM = (6.67 · 10-1). (1400) (5.97219-108²4) ~ 5.57·1017
Work done lifting from Earth surface (6,371,000 m)
X²
X²
x²
(units are kg. 5²,
↑
to orbiting altitude (6,37) km +850 km = 7,221 km = 7,221,000 m)
M
sec²
a.k.a. Newtons)
is
praz
1272231-1046 5.57.1080
16.371.106
x²
=
5.57.105
6.371-106
7221-106
= (5,57-102²) - x 22²221 10 = 5,57 - 108² x 16.5715186
[X-176.371-·106
$6.371.106
7.221-186
5.57.10¹17
7.221.10⁰
10" (5.57 - $57) ≈ 1.03. 10¹" N´m, aka
5.57
=
~
6.371
[7.03×10⁰ Joules
Transcribed Image Text:(238) F(x) = G₁MM = (6.67 · 10-1). (1400) (5.97219-108²4) ~ 5.57·1017 Work done lifting from Earth surface (6,371,000 m) X² X² x² (units are kg. 5², ↑ to orbiting altitude (6,37) km +850 km = 7,221 km = 7,221,000 m) M sec² a.k.a. Newtons) is praz 1272231-1046 5.57.1080 16.371.106 x² = 5.57.105 6.371-106 7221-106 = (5,57-102²) - x 22²221 10 = 5,57 - 108² x 16.5715186 [X-176.371-·106 $6.371.106 7.221-186 5.57.10¹17 7.221.10⁰ 10" (5.57 - $57) ≈ 1.03. 10¹" N´m, aka 5.57 = ~ 6.371 [7.03×10⁰ Joules
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