This is a direct calculation question. You are in a spacecraft orbiting Jupiter. The mass of Jupiter is 1.90E+27 kg, and it's radius is 7.14E+4 km. What is the escape velocity from the surface? m/s. At a height of 7 radii above the surface of the Jupiter, what is the circular orbital velocity (orbital velocity, not escape velocity)? m/s Note: If your answer requires scientific notation, remember that OWL uses "e" notation: 1.1 x 105 is 1.1e5 to OWL.
This is a direct calculation question. You are in a spacecraft orbiting Jupiter. The mass of Jupiter is 1.90E+27 kg, and it's radius is 7.14E+4 km. What is the escape velocity from the surface? m/s. At a height of 7 radii above the surface of the Jupiter, what is the circular orbital velocity (orbital velocity, not escape velocity)? m/s Note: If your answer requires scientific notation, remember that OWL uses "e" notation: 1.1 x 105 is 1.1e5 to OWL.
Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Chapter1: The Study Of Minerals
Section: Chapter Questions
Problem 1LR
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Transcribed Image Text:You may attempt this question 2 more times for credit.
This is a direct calculation question. You are in a spacecraft orbiting Jupiter.
The mass of Jupiter is 1.90E+27 kg, and it's radius is 7.14E+4 km.
What is the escape velocity from the surface?
m/s.
At a height of 7 radii above the surface of the Jupiter, what is the circular orbital velocity (orbital velocity, not escape velocity)?
m/s
Note: If your answer requires scientific notation, remember that OWL uses "e" notation: 1.1 x 105 is 1.1e5 to OWL.
Expert Solution
Step 1 Answer
Answer to the first question
Given the mass of Jupiter(M)=1.90×1027kg
Radius(r)= 1.14×104km, G= 6.67×10-11m3Kg-1s-2
Escape velocity (Vs)= ✓(2GM/r)
Vs= ✓2×6.67×10-11 m3 Kg -1s-1× 1.90×1027kg/ 1.14×104km =✓(22.133×1012)km/s
Vs=4.704×106km/sec
Answer to the second question
Given the mass of Jupiter(M)=1.90×1027kg
The space craft orbiting Jupiter is 7 radii above the surface of the Jupiter so the radii here radius is the addition of the height and the radius of the jupiter=
7×1.44×104km+1.44×104km = 8×1.44×104km
G= 6.67×10-11m3Kg-1s-2
Orbital velocity (vo)=✓(GM/r)
V0=✓(6.67×10 -11m -3Kg -1s-2 ×1.90×1027/8×1.44×104) = ✓1.100×1012
V0=0.316×106 m/s
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