急~~~~~有關普通物理的問題!我想了超久!還是想不出來!
第一題敘述:A satellite used in a cellular telephone network has a mass of 1870kg and is in a circular orbit at a height of 790km above the surface of the earth.
問:Part B
What fraction is this of the satellite's weight at the surface of the earth?
Take the free-fall acceleration at the surface of the earth to be g= 9.80m/s2 .
我Part A會算Fgrav=1.45*10000 N ps(G=6.67×10的−11次方,me= 5.97×10的24次方 kg,re= 6.38×10的6次方 m)
第二題敘述:The Navstar Global Positioning System (GPS) utilizes a group of 24 satellites orbiting the Earth. Using "triangulation" and signals transmitted by these satellites, the position of a receiver on the Earth can be determined to within an accuracy of a few centimeters. The satellite orbits are distributed evenly around the Earth, with four satellites in each of six orbits, allowing continuous navigational "fixes." The satellites orbit at an altitude of approximately 11000 nautical miles [1 nautical mile = 1.852km = 6076ft].
Part A
Determine the speed of each satellite.
Express your answer using two significant figures.
v=?m/s
Part B
Determine the period of each satellite.
Express your answer using two significant figures.
T=?h
謝謝你的辛苦!
ps:我沒答案,麻煩告訴我怎麼算和計算過程!
急~有關普通物理題目的問題!
2009-10-17 11:15 pm
回答 (1)
2009-10-18 3:05 am
✔ 最佳答案
1. The weight obeys inverse square lawSo, mg'/mg = (R/R')^2
= [(6.36 X 10^6) / (6.36 X 10^6 + 790 X 10^3)]^2
= 0.791
2.a. By GM / R^2 = v^2 / R
v = sqrt[GM/R] = sqrt[(6.67 X 10^-11)(5.98 X 10^24) / (6.36 X 10^6 + 11000 X 1.852 X 10^3)]
= 3.86 X 10^3 m/s
b. By T = 2pir / v
= 2pi(6.36 X 10^6 + 11000 X 1.852 X 10^3) / (3.86 X 10^3)
= 4.35 X 10^4 s
= 12.08 h
參考: Physics king
收錄日期: 2021-04-19 15:59:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091017000015KK04837