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An Attempt to Correct the Formula of Universal Gravitation Based on the Proportion of Gravitons Emitted from the Planet to the Outside of the Sphere—The Ratio of Gravitons Outside of the Sphere in Deflection Gravity Theory
DOI: 10.12677/AAS.2023.113003, PDF, HTML, XML, 下载: 159  浏览: 291

Abstract: Based on the theory of deflected gravity, this article calculates the number of gravitons sent from the planet’s nuclei to the outside of the sphere by layering the planet. It shows that the number of gravitons sent from the planet’s nuclei to the outside of the sphere is less than one thousandth of the planet’s radius. The layered calculation shows that for a planet with a relatively large radius, the number of gravitons sent to the outside of the sphere is only related to the radius of the planet. The formula of universal gravitation is then modified to show that the gravity between planets is proportional to the square of the radius of the planet, inversely proportional to the square of the dis-tance from the planet.

1. 偏转引力理论简述

1687年牛顿发现了万有引力 [1] ，1915年爱因斯坦在广义相对论 [2] [3] 中把引力等效成时空的弯曲，偏转引力理论 [4] [5] [6] 从微观作用机理上揭示了引力的作用过程。

Figure 1. Microphotons and single photons

${\lambda }_{0}=1.6×{10}^{-15}\text{\hspace{0.17em}}\text{m}$ (1)

${f}_{0}=1.875×{10}^{23}\text{\hspace{0.17em}}\text{Hz}$ (2)

${T}_{0}=5.33×{10}^{-24}\text{\hspace{0.17em}}\text{s}$ (3)

${k}_{ng}=0.682$ (4)

${n}_{ng}=6.318×{10}^{21}$ (5)

2. 星球发往球外引力子数量

Figure 2. Planet layering

${\rho }_{s}=\frac{{m}_{s}}{\frac{4}{3}\pi {r}_{s}^{3}}=\frac{\text{3}{m}_{s}}{\text{4}\pi {r}_{s}^{3}}$ (6)

${m}_{s}=\frac{4}{3}\pi {r}_{s}^{3}{\rho }_{s}$ (7)

${r}_{s}={\left(\frac{\text{3}{m}_{s}}{\text{4}\pi {\rho }_{s}}\right)}^{1/3}$ (8)

$\frac{\frac{4}{3}\pi {r}_{s}^{3}}{\frac{{m}_{s}}{{m}_{0}}}=\frac{4}{3}\pi {r}_{e}^{3}$ (9)

${r}_{e}={r}_{s}{\left(\frac{{m}_{0}}{{m}_{s}}\right)}^{1/3}={r}_{s}{\left(\frac{{m}_{0}}{\frac{4}{3}\pi {r}_{s}^{3}{\rho }_{s}}\right)}^{1/3}={\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{1/3}$ (10)

${N}_{s}=\frac{{r}_{s}}{{r}_{e}}=\frac{{r}_{s}}{{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{1/3}}={r}_{s}{\left(\frac{4\pi {\rho }_{s}}{3{m}_{0}}\right)}^{1/3}$ (11)

${k}_{s1}=\frac{\frac{4\pi {\left({r}_{i}+2{r}_{e}\right)}^{2}}{\pi {r}_{e}^{2}}\pi {r}_{0}^{2}{k}_{ng}}{4\pi {\left({r}_{i}+2{r}_{e}\right)}^{2}}={k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}$ (12)

ri + 4re层所有核子有效面积占当层总球面的比例为：

${k}_{s2}=\frac{\frac{\left(4\pi -{k}_{s1}\right){\left({r}_{i}+4{r}_{e}\right)}^{2}}{\pi {r}_{e}^{2}}\pi {r}_{0}^{2}{k}_{ng}}{\left(4\pi -{k}_{s1}\right){\left({r}_{i}+4{r}_{e}\right)}^{2}}={k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}$ (13)

${N}_{so}=\frac{4\pi }{{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}}=\frac{4\pi {r}_{e}^{2}}{{k}_{ng}{r}_{0}^{2}}=\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}{\left(\frac{\text{3}{m}_{0}}{\text{4}\pi {\rho }_{s}}\right)}^{2/3}$ (14)

${N}_{si}={N}_{s}-{N}_{so}$ (15)

${r}_{so}={N}_{so}×2{r}_{e}=\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{2/3}×2{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{1/3}=\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}=\frac{{k}_{s\rho }}{{\rho }_{s}}=\frac{22956}{{\rho }_{s}}$ (16)

$\begin{array}{c}{n}_{go}=\frac{4\pi {\left({r}_{i}+2{r}_{e}\right)}^{2}}{\pi {r}_{e}^{2}}×{n}_{ng}×{k}_{o}×{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}+\frac{4\pi {\left({r}_{i}+4{r}_{e}\right)}^{2}}{\pi {r}_{e}^{2}}×{n}_{ng}×{k}_{o}×2{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{4\pi {\left({r}_{i}+6{r}_{e}\right)}^{2}}{\pi {r}_{e}^{2}}×{n}_{ng}×{k}_{o}×3{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}+\frac{4\pi {\left({r}_{i}+8{r}_{e}\right)}^{2}}{\pi {r}_{e}^{2}}×{n}_{ng}×{k}_{o}×4{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}+\cdots \end{array}$ (17)

${n}_{go}=\frac{4\pi }{\pi {r}_{e}^{2}}{n}_{n\text{g}}{k}_{o}{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}\left[{\left({r}_{i}+2{r}_{e}\right)}^{2}+\text{2}{\left({r}_{i}+4{r}_{e}\right)}^{2}+\text{3}{\left({r}_{i}+6{r}_{e}\right)}^{2}+\text{4}{\left({r}_{i}+8{r}_{e}\right)}^{2}+\cdots \right]$ (18)

$\begin{array}{c}{n}_{go}=\frac{4{n}_{n\text{g}}{k}_{o}{k}_{ng}{r}_{0}^{2}}{{r}_{e}^{\text{4}}}\left[{r}_{i}^{2}+2×2{r}_{i}{r}_{e}+{2}^{2}{r}_{e}^{2}+\text{2}{r}_{i}^{2}+\text{2}×\text{2}×4{r}_{i}{r}_{e}+\text{2}×{4}^{2}{r}_{e}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{3}{r}_{i}^{2}+\text{3}×2×6{r}_{i}{r}_{e}+\text{3}×{6}^{2}{r}_{e}^{2}+\text{4}{r}_{i}^{2}+\text{4}×2×8{r}_{i}{r}_{e}+\text{4}×{8}^{2}{r}_{e}^{2}+\cdots \right]\end{array}$ (19)

$\begin{array}{c}{n}_{go}=\frac{4{n}_{n\text{g}}{k}_{o}{k}_{ng}{r}_{0}^{2}}{{r}_{e}^{\text{4}}}\left[{r}_{i}^{2}+\text{2}{r}_{i}^{2}+\text{3}{r}_{i}^{2}+\text{4}{r}_{i}^{2}+\cdots +2×2{r}_{i}{r}_{e}+\text{2}×\text{2}×4{r}_{i}{r}_{e}+\text{3}×2×6{r}_{i}{r}_{e}+\text{4}×2×8{r}_{i}{r}_{e}+\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{2}^{2}{r}_{e}^{2}+\text{2}×{4}^{2}{r}_{e}^{2}+\text{3}×{6}^{2}{r}_{e}^{2}+\text{4}×{8}^{2}{r}_{e}^{2}+\cdots \right]\end{array}$ (20)

${n}_{go}=\frac{4{n}_{n\text{g}}{k}_{o}{k}_{ng}{r}_{0}^{2}}{{r}_{e}^{\text{4}}}\left[{r}_{i}^{2}\left(\text{1}+\text{2}+\text{3}+\text{4}+\cdots \right)+\text{4}{r}_{i}{r}_{e}\left(\text{1}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}+{\text{4}}^{\text{2}}+\cdots \right)+{2}^{2}{r}_{e}^{2}\left(\text{1}+{\text{2}}^{\text{3}}+{\text{3}}^{\text{3}}+{\text{4}}^{\text{3}}+\cdots \right)\right]$ (21)

${n}_{go}=\frac{4{n}_{ng}{k}_{o}{k}_{ng}{r}_{0}^{2}}{{r}_{e}^{4}}\left[{r}_{i}^{2}\frac{{N}_{so}\left({N}_{so}+1\right)}{2}+4{r}_{i}{r}_{e}\frac{{N}_{so}\left({N}_{so}+1\right)\left(2{N}_{so}+1\right)}{6}+{2}^{2}{r}_{e}^{2}{\left(\frac{{N}_{so}\left({N}_{so}+1\right)}{2}\right)}^{2}\right]$ (22)

${n}_{go}\approx \frac{4{n}_{n\text{g}}{k}_{o}{k}_{ng}{r}_{0}^{2}}{{r}_{e}^{\text{4}}}\left[\frac{1}{\text{2}}{r}_{i}^{2}{N}_{so}^{2}+\frac{4}{\text{3}}{r}_{i}{r}_{e}{N}_{so}^{3}+{r}_{e}^{2}{N}_{so}^{4}\right]$ (23)

${n}_{go}=\frac{4{n}_{ng}{k}_{o}{k}_{ng}{r}_{0}^{2}}{{r}_{e}^{4}}{N}_{so}^{2}\left[\frac{1}{2}{\left({r}_{s}-{r}_{so}\right)}^{2}+\frac{4}{3}\left({r}_{s}-{r}_{so}\right){r}_{e}{N}_{so}+{r}_{e}^{2}{N}_{so}^{2}\right]$ (24)

$\begin{array}{c}{n}_{go}=\frac{4{n}_{ng}{k}_{o}{k}_{ng}{r}_{0}^{2}}{{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{4/3}}{\left(\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}\right)}^{2}{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{4/3}\left[\frac{1}{2}{\left({r}_{s}-\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)}^{2}+\frac{4}{3}\left({r}_{s}-\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right){\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{1/3}\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{2/3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{2/3}{\left(\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}\right)}^{2}{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{4/3}\right]\end{array}$ (25)

${n}_{go}=4{n}_{ng}{k}_{o}{k}_{ng}{r}_{0}^{2}{\left(\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}\right)}^{2}\left[\frac{1}{2}{\left({r}_{s}-\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)}^{2}+\frac{4}{3}\left({r}_{s}-\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}+{\left(\frac{3{m}_{0}}{4\pi {\rho }_{s}}\right)}^{2}{\left(\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}\right)}^{2}\right]$ (26)

${n}_{go}=4{n}_{ng}{k}_{o}{k}_{ng}{r}_{0}^{2}{\left(\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}\right)}^{2}\left[\frac{1}{2}{r}_{s}^{2}-\frac{1}{2}2{r}_{s}\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}+\frac{1}{2}{\left(\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)}^{2}+{r}_{s}\frac{4{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}-\frac{6{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\frac{4{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}+{\left(\frac{3{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)}^{2}\right]$ (27)

${n}_{go}=2{n}_{n0}{k}_{o}{k}_{ng}{r}_{0}^{2}{\left(\frac{4\pi }{{k}_{ng}{r}_{0}^{2}}\right)}^{2}\left[{r}_{s}^{2}-12{r}_{s}\frac{{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}+8{r}_{s}\frac{{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}+36{\left(\frac{{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)}^{2}-48{\left(\frac{{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)}^{2}+18{\left(\frac{{m}_{0}}{{k}_{ng}{r}_{0}^{2}{\rho }_{s}}\right)}^{2}\right]$ (28)

${n}_{go}=\frac{32{\pi }^{2}{n}_{ng}{k}_{o}}{{k}_{ng}{r}_{0}^{2}}\left[{r}_{s}^{2}-\frac{4{m}_{0}}{{k}_{ng}{r}_{0}^{2}}\frac{{r}_{s}}{{\rho }_{s}}+6{\left(\frac{{m}_{0}}{{k}_{ng}{r}_{0}^{2}}\right)}^{2}\frac{1}{{\rho }_{s}^{2}}\right]$ (29)

${n}_{go}=\text{2}\text{.514}×{\text{10}}^{\text{54}}\left[{r}_{s}^{2}-1.53×{\text{10}}^{4}\frac{{r}_{s}}{{\rho }_{s}}+8.783×{\text{10}}^{7}\frac{\text{1}}{{\rho }_{s}^{2}}\right]$ (30)

${k}_{1}=\frac{{n}_{go}}{\frac{{m}_{s}}{{m}_{0}}{n}_{ng}}=\frac{{n}_{go}{m}_{0}}{{n}_{ng}{m}_{s}}$ (31)

3. 地球与太阳球外引力子数量

3.1. 地球球外引力子数量

${r}_{s3}=1.7×{\text{10}}^{4}\text{\hspace{0.17em}}\text{m}$ (32)

${\rho }_{s3}=2800\text{\hspace{0.17em}}\text{kg}/{\text{m}}^{\text{3}}$ (33)

${r}_{s}=\text{6}\text{.371}×{\text{10}}^{\text{6}}\text{\hspace{0.17em}}\text{m}$ (34)

$\begin{array}{c}{m}_{s3}=\frac{4}{3}\pi \left[{r}_{s}^{3}-{\left({r}_{s}-{r}_{s3}\right)}^{3}\right]{\rho }_{s3}\\ =\frac{4}{3}×3.14\left[{\left(6.371×{10}^{6}\right)}^{3}-{\left(6.371×{10}^{6}-1.7×{10}^{4}\right)}^{3}\right]×2800\\ =2.421×{10}^{22}\end{array}$ (35)

$\frac{4}{3}\pi {r}_{e3}^{3}=\frac{\frac{4}{3}\pi \left[{r}_{s}^{3}-{\left({r}_{s}-{r}_{s3}\right)}^{3}\right]}{\frac{{m}_{s3}}{{m}_{0}}}$ (36)

$\begin{array}{c}{r}_{e3}=\sqrt[3]{\frac{{m}_{0}}{{m}_{s3}}\left[{r}_{s}^{3}-{\left({r}_{s}-{r}_{s3}\right)}^{3}\right]}\\ =\sqrt[3]{\frac{\text{1}\text{.67}×{\text{10}}^{-27}}{2.478×{\text{10}}^{22}}\left[{\left(\text{6}\text{.371}×{\text{10}}^{\text{6}}\right)}^{\text{3}}-{\left(\text{6}\text{.371}×{\text{10}}^{\text{6}}-1.7×{\text{10}}^{4}\right)}^{3}\right]}\\ =5.222×{\text{10}}^{-11}\end{array}$ (37)

${N}_{s3}=\frac{{r}_{s3}}{{r}_{e3}}=\frac{1.7×{\text{10}}^{4}}{5.222×{\text{10}}^{-11}}=3.256×{\text{10}}^{14}$ (38)

${N}_{so3}=\frac{4\pi }{{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}}=\frac{4\pi {r}_{e3}^{2}}{{k}_{ng}{r}_{0}^{2}}=\frac{4×3.14×{\left(5.222×{10}^{-11}\right)}^{2}}{0.682×{\left(8×{10}^{-16}\right)}^{2}}=7.850×{10}^{10}$ (39)

${n}_{s}{}_{\text{3}}>{n}_{so}{}_{\text{3}}$ ，说明发射到球外引力子的核子全部由地壳层提供。引力子能够发射到球外的核子厚度为：

${r}_{so}={N}_{so\text{3}}×2{r}_{e3}=7.850×{\text{10}}^{10}×2×5.222×{\text{10}}^{-11}=\text{8}\text{.199}\text{\hspace{0.17em}}\text{m}$ (40)

$\begin{array}{c}{n}_{so}=2.514×{10}^{54}\left[{r}_{s}{}^{2}-1.53×{10}^{4}\frac{{r}_{s}}{{\rho }_{s}}+8.783×{10}^{7}\frac{1}{{\rho }_{s}^{2}}\right]\\ =2.514×{10}^{54}\left[{\left(6.371×{10}^{6}\right)}^{2}-1.53×{10}^{4}\frac{6.371×{10}^{6}}{2800}+8.783×{10}^{7}\frac{1}{{2800}^{2}}\right]\\ =1.020×{10}^{68}\end{array}$ (41)

3.2. 太阳球外引力子数量

Figure 3. Earth structure

Figure 4. Solar structure

${r}_{s3}=\text{5}×{\text{10}}^{\text{5}}\text{\hspace{0.17em}}\text{m}$ (42)

${\rho }_{s3}=\text{1}0\text{\hspace{0.17em}}\text{kg}/{\text{m}}^{\text{3}}$ (43)

${r}_{s}=\text{6}\text{.96}×{\text{10}}^{\text{8}}\text{\hspace{0.17em}}\text{m}$ (44)

$\begin{array}{c}{m}_{s3}=\frac{4}{3}\pi \left[{r}_{s}^{3}-{\left({r}_{s}-{r}_{s3}\right)}^{3}\right]{\rho }_{s3}\\ =\frac{4}{3}×\text{3}\text{.14}\left[{\left(\text{6}\text{.96}×{\text{10}}^{\text{8}}\right)}^{\text{3}}-{\left(\text{6}\text{.96}×{\text{10}}^{\text{8}}-\text{5}×{\text{10}}^{\text{5}}\right)}^{3}\right]×\text{10}\\ =3.041×{\text{10}}^{25}\end{array}$ (45)

$\frac{4}{3}\pi {r}_{e3}^{3}=\frac{\frac{4}{3}\pi \left[{r}_{s}^{3}-{\left({r}_{s}-{r}_{s3}\right)}^{3}\right]}{\frac{{m}_{s3}}{{m}_{0}}}$ (46)

$\begin{array}{c}{r}_{e3}=\sqrt[3]{\frac{{m}_{0}}{{m}_{s3}}\left[{r}_{s}^{3}-{\left({r}_{s}-{r}_{s3}\right)}^{3}\right]}\\ =\sqrt[3]{\frac{1.67×{10}^{-27}}{3.041×{10}^{25}}\left[{\left(6.96×{10}^{8}\right)}^{3}-{\left(6.96×{10}^{8}-5×{10}^{5}\right)}^{3}\right]}\\ =3.416×{10}^{-10}\end{array}$ (47)

${N}_{s3}=\frac{{r}_{s3}}{{r}_{e3}}=\frac{\text{5}×{\text{10}}^{\text{5}}}{\text{3}\text{.416}×{\text{10}}^{-10}}=1.464×{\text{10}}^{15}$ (48)

${N}_{so3}=\frac{4\pi }{{k}_{ng}\frac{{r}_{0}^{2}}{{r}_{e}^{2}}}=\frac{4\pi {r}_{e3}^{2}}{{k}_{ng}{r}_{0}^{2}}=\frac{4×3.14×{\left(3.416×{10}^{-10}\right)}^{2}}{0.682×{\left(8×{10}^{-16}\right)}^{2}}=3.360×{10}^{12}$ (49)

${n}_{s3}>{n}_{so3}$ ，说明发射到球外引力子的核子全部由光球层提供。引力子能够发射到球外的核子层厚度为：

${r}_{so}={N}_{so\text{3}}×2{r}_{e3}=\text{3}\text{.360}×{\text{10}}^{1\text{2}}×2×\text{3}\text{.416}×{\text{10}}^{-10}=\text{2}296\text{\hspace{0.17em}}\text{m}$ (50)

${n}_{so}=\text{2}\text{.514}×{\text{10}}^{54}\left[{\left(6.96×{\text{10}}^{8}\right)}^{\text{2}}-1.53×{\text{10}}^{4}×\frac{6.96×{\text{10}}^{8}}{10}+8.783×{\text{10}}^{7}×\frac{\text{1}}{{10}^{\text{2}}}\right]=\text{1}\text{.218}×{\text{10}}^{72}$ (51)

3.3. 球外引力子数量公式30的简化

Table 1. The number of gravitons emitted by the planets in the solar system to the outside world

${n}_{go}\approx {k}_{gr}{r}_{s}^{2}=\text{2}\text{.514}×{\text{10}}^{54}{r}_{s}^{2}$ (52)

4. 对万有引力的修正

$F=G\frac{{m}_{1}{m}_{2}}{{R}^{2}}$ (53)

$F={G}_{1}\frac{\frac{{m}_{1}}{{m}_{0}}{k}_{1}\frac{{m}_{2}}{{m}_{0}}{k}_{2}}{{R}^{2}}$ (54)

$F={G}_{1}\frac{{n}_{go1}{n}_{go2}}{{R}^{2}}$ (55)

$F={G}_{1}\frac{{k}_{gr}{r}_{s\text{1}}^{2}{k}_{gr}{r}_{s\text{2}}^{2}}{{R}^{2}}$ (56)

$F={G}_{r}\frac{{r}_{s\text{1}}^{2}{r}_{s\text{2}}^{2}}{{R}^{2}}$ (57)

$F={G}_{S}\frac{{S}_{1}{S}_{2}}{{R}^{2}}$ (58)

5. 结论

nng = 6.318 × 1021

nng = 6.318 × 1021

${}_{1}{}^{2}\text{H}$ 的结合能为2.224 MeV

k = 22,956

k = 22,956

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