Asked by Leanna
1. At what height above the earth is the free-fall acceleration 30% of its value at the surface?
2. What is the speed of a satellite orbiting at that height?
2. What is the speed of a satellite orbiting at that height?
Answers
Answered by
Elena
1.
g =GM/R²
g₁=GM/(R+h)²
g₁/g=30/100= GM R²/(R+h)²GM =>
0.3= R²/(R+h)²
Solve for “h” (Earth’s radius is R = 6.378•10⁶ m)
2.
mv²/(R+h)= GmM/(R+h)²
v=sqrt { GM/(R+h)}
g =GM/R²
g₁=GM/(R+h)²
g₁/g=30/100= GM R²/(R+h)²GM =>
0.3= R²/(R+h)²
Solve for “h” (Earth’s radius is R = 6.378•10⁶ m)
2.
mv²/(R+h)= GmM/(R+h)²
v=sqrt { GM/(R+h)}
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.