某衛星繞地球作圓周運動之週期為T 軌道半徑r

2013-02-07 9:11 am

某衛星繞地球作圓周運動之週期為T 軌道半徑r 則

1.地球質量為何

2.衛星軌道之重力場強度為何

3.若地球之半徑為R 則地表重力場強度為何

4.地球的密度為何

5.衛星運行速率為何

更新1:

答案是 1. 4拍平方R3次方/GT平方 2. 4拍平方r/T平方 3. 4拍平方r3次方/R平方T平方 4. 3拍r3次方/GR3次方T平方 5. 2拍r/T 1. g=GM/r平方 =4拍平方r/T平方 M=4拍平方R3次方/GT平方 2. g=a=4拍平方r/T平方 3. g=GM/R平方 =G*(4拍平方r3次方)/GT平方/R平方 =4拍平方r3次方/R平方T平方 4. g=4/3拍DGR=4拍平方r3次方/R平方T平方 D=3拍r3次方/GR3次方T平方

更新2:

5.等速率圓周運動v=2拍r/T

回答 (1)

天同

2013-02-07 4:22 pm

✔ 最佳答案

1. Gravitational force on satellite = GMm/r^2
where G is the Universal Gravitational Constant
M is the mass of earth
m is the mass of satellite

Centripetal force required = mr.(2.pi/T)^2
Hence, GMm/r^2 = m.r(4.pi^2)/T^2
M = (4.pi^2)r^3/(GT^2)

2. Acceleration due to gravity at orbit
= GM/r^2
= G[(4.pi^2)r^3/(GT^2)]/r^2
= (4.pi^2)r/T^2

3. Acceleration due to gravity on earth surface g = GM/R^2
i.e. g = G[(4.pi^2)r^3/(GT^2)]/R^2
g = (4.pi^2)r^3/(TR)^2

4. Volume of earth = (4.pi/3)R^3
Since earht mass = (4.pi^2)r^3/(GT^2)
density of earth = [(4.pi^2)r^3/(GT^2)]/[ (4.pi/3)R^3]
= 3.pi.r^3/(GT^2.R^3)

5. Speed of satellite = (2.pi.r)/T

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