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兩個(gè)高中物理推論的數(shù)學(xué)證明

2023-07-11 12:00 作者:げいしも_蕓  | 我要投稿

推論一:

對(duì)于質(zhì)地均勻且密度為ρ?,半徑為R的球,其對(duì)球外一質(zhì)點(diǎn)的引力等價(jià)于在球心處質(zhì)量為M(M=4πR3ρ?/3)的質(zhì)點(diǎn)對(duì)該點(diǎn)的引力

證明:

設(shè)該球所占有的空間為:

%5COmega_1%3D%5C%7B(x%2Cy%2Cz)%5Cvert%20x%5E2%2By%5E2%2Bz%5E2%5Cleq%20R%5E2%5C%7D

球外一質(zhì)點(diǎn)

M_1(0%2C0%2Ca)(a%3ER)

由對(duì)稱(chēng)性知其引力在x,y軸上的分量為0,則有:

F_z%3D%5Ciiint%5Climits_%7B%5COmega_1%7D%5Cfrac%7BmG%5Crho_0(z-a)%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%5Cfrac%2032%7D%5Ctext%20dv

%3DGm%5Crho_0%5Cint_%7B-R%7D%5ER(z-a)%5Ctext%20dz%5Ciint%5Climits_%7Bx%5E2%2By%5E2%5Cleq%20R%5E2-z%5E2%7D%5Cfrac%7B%5Ctext%20dx%20%5Ctext%20dy%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%7B%5Cfrac%2032%7D%7D

對(duì)二重積分部分使用極坐標(biāo)換元,得到:

F_z%3DGm%5Crho_0%5Cint_%7B-R%7D%5E%7BR%7D(z-a)%5Ctext%20dz%5Cint_0%5E%7B2%5Cpi%7D%5Ctext%20d%5Ctheta%5Cint_0%5E%7B%5Csqrt%7BR%5E2-z%5E2%7D%7D%5Cfrac%7B%5Crho%5Ctext%20d%5Crho%7D%7B%5B%5Crho%5E2%2B(z-a)%5E2%5D%5E%5Cfrac%2032%7D

%3D2%5Cpi%20Gm%5Crho_0%5Cint_%7B-R%7D%5E%7BR%7D(z-a)(%5Cfrac%7B1%7D%7Ba-z%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D)%5Ctext%20dz

%3D2%5Cpi%20Gm%5Crho_0%5Cleft%20(%20-2R%2B2R-%5Cfrac%7B2R%5E3%7D%7B3a%5E2%7D%5Cright)

%3D-Gm%5Ccdot%5Cfrac43%5Cpi%20R%5E3%5Crho_0%5Ccdot%20%5Cfrac%201%7Ba%5E2%7D%3D-%5Cfrac%7BGMm%7D%7Ba%5E2%7D

推論一得證.

推論二:

對(duì)于一外半徑為R,內(nèi)半徑為r(R≥r),質(zhì)地均勻的空心球體內(nèi)一點(diǎn),該空心球體對(duì)該質(zhì)點(diǎn)的引力為0

證明:

設(shè)該物體所占有的空間為:

%5COmega_2%3D%5C%7B(x%2Cy%2Cz)%5Cvert%20r%5E2%5Cleq%20x%5E2%2By%5E2%2Bz%5E2%5Cleq%20R%5E2%5C%7D(R%5Cgeq%20r)

內(nèi)部一質(zhì)點(diǎn):

M_2(0%2C0%2Ca)(0%5Cleq%20a%20%5Cleq%20r)

同樣由對(duì)稱(chēng)性可以知道引力在x,y方向上的分量為0,于是有:

F_z%3D%5Ciiint%5Climits_%7B%5COmega_2%7D%5Cfrac%7BmG%5Crho_0(z-a)%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%7B%5Cfrac%2032%7D%7D%5Ctext%20dv

%3DmG%5Crho_0(I_1%2BI_2%2BI_3)

其中,

I_1%3D%5Cint_%7B-R%7D%5E%7B-r%7D(z-a)%5Ctext%20dz%5Ciint%5Climits_%7Bx%5E2%2By%5E2%5Cleq%20R%5E2-z%5E2%7D%5Cfrac%7B%5Ctext%20dx%5Ctext%20dy%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%5Cfrac23%7D

I_2%3D%5Cint_%7B-r%7D%5E%7Br%7D(z-a)%5Ctext%20dz%5Ciint%5Climits_%7Br%5E2-z%5E2%5Cleq%20x%5E2%2By%5E2%5Cleq%20R%5E2-z%5E2%7D%5Cfrac%7B%5Ctext%20dx%5Ctext%20dy%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%5Cfrac23%7D

I_3%3D%5Cint_%7Br%7D%5E%7BR%7D(z-a)%5Ctext%20dz%5Ciint%5Climits_%7Bx%5E2%2By%5E2%5Cleq%20R%5E2-z%5E2%7D%5Cfrac%7B%5Ctext%20dx%5Ctext%20dy%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%5Cfrac23%7D

現(xiàn)對(duì)后面的三個(gè)二重積分化簡(jiǎn)

當(dāng)-R≤z≤-r時(shí)

%5Ciint%5Climits_%7Bx%5E2%2By%5E2%5Cleq%20R%5E2-z%5E2%7D%5Cfrac%7B%5Ctext%20dx%5Ctext%20dy%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%5Cfrac23%7D

%3D%5Cint_0%5E%7B2%5Cpi%7D%5Ctext%20d%5Ctheta%5Cint_0%5E%7B%5Csqrt%7BR%5E2-z%5E2%7D%7D%5Cfrac%7B%5Crho%20%5Ctext%20d%5Crho%7D%7B%5B%5Crho%5E2%2B(z-a)%5E2%5D%5E%5Cfrac%2032%7D%3D2%5Cpi%5Cleft%20(%5Cfrac%7B1%7D%7Ba-z%7D-%5Cfrac%201%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D%5Cright%20)

當(dāng)-r≤z≤r時(shí)

%5Ciint%5Climits_%7Br%5E2-z%5E2%5Cleq%20x%5E2%2By%5E2%5Cleq%20R%5E2-z%5E2%7D%5Cfrac%7B%5Ctext%20dx%5Ctext%20dy%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%5Cfrac%2032%7D

%3D%5Cint_0%5E%7B2%5Cpi%7D%5Ctext%20d%5Ctheta%5Cint_%7B%5Csqrt%7Br%5E2-z%5E2%7D%7D%5E%7B%5Csqrt%7BR%5E2-z%5E2%7D%7D%5Cfrac%7B%5Crho%20%5Ctext%20d%5Crho%7D%7B%5B%5Crho%5E2%2B(z-a)%5E2%5D%5E%5Cfrac%2032%7D%3D2%5Cpi%5Cleft%20(%5Cfrac%7B1%7D%7B%5Csqrt%7Br%5E2-2az%2Ba%5E2%7D%7D-%5Cfrac%201%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D%5Cright%20)

當(dāng)r≤z≤R時(shí)

%5Ciint%5Climits_%7Bx%5E2%2By%5E2%5Cleq%20R%5E2-z%5E2%7D%5Cfrac%7B%5Ctext%20dx%5Ctext%20dy%7D%7B%5Bx%5E2%2By%5E2%2B(z-a)%5E2%5D%5E%5Cfrac%2032%7D

%3D%5Cint_0%5E%7B2%5Cpi%7D%5Ctext%20d%5Ctheta%5Cint_0%5E%7B%5Csqrt%7BR%5E2-z%5E2%7D%7D%5Cfrac%7B%5Crho%20%5Ctext%20d%5Crho%7D%7B%5B%5Crho%5E2%2B(z-a)%5E2%5D%5E%5Cfrac%2032%7D%3D2%5Cpi%5Cleft%20(%5Cfrac%7B1%7D%7Bz-a%7D-%5Cfrac%201%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D%5Cright%20)

代回原積分表達(dá)式得:

I_1%3D2%5Cpi%5Cint_%7B-R%7D%5E%7B-r%7D%5Cleft%20(%20-1-%5Cfrac%20%7Bz-a%7D%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D%5Cright%20)%20%5Ctext%20dz

I_2%3D2%5Cpi%5Cint_%7B-r%7D%5E%7Br%7D%5Cleft%20(%20%5Cfrac%7Bz-a%7D%7B%5Csqrt%7Br%5E2-2az%2Ba%5E2%7D%7D-%5Cfrac%20%7Bz-a%7D%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D%5Cright%20)%20%5Ctext%20dz

I_3%3D2%5Cpi%5Cint_%7Br%7D%5E%7BR%7D%5Cleft%20(%201-%5Cfrac%20%7Bz-a%7D%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D%5Cright%20)%20%5Ctext%20dz

于是得到:

F_z%3DmG%5Crho_0(I_1%2BI_2%2BI_2)

%3D2m%5Cpi%20G%5Crho_0%5Cleft%20(%5Cint_%7B-r%7D%5Er%5Cfrac%7Bz-a%7D%7B%5Csqrt%7Br%5E2-2az%2Ba%5E2%7D%7D%5Ctext%20dz-%5Cint_%7B-R%7D%5ER%20%5Cfrac%7Bz-a%7D%7B%5Csqrt%7BR%5E2-2az%2Ba%5E2%7D%7D%5Ctext%20dz%5Cright%20)

%3D0

推論二得證.

兩個(gè)高中物理推論的數(shù)學(xué)證明的評(píng)論 (共 條)

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