Equation (2-30) is an approximation correct only if the gravitational time-dilation effect is small. In this exercise, it is also assumed to be small. but we still allow for a nonuniform gravitational field. We start with (2-29), based on the Doppler effect in the accelerating frame. Consider two elevations, the lower at r₁ and the upper at ${\mathbf{r}}_{\mathbf{1}}\mathbf{+}\mathbf{dr}$. Equation (2·29) becomes

$\frac{\mathbf{f}\left(r_1+\mathrm{dr}\right)}{\mathbf{f}\left(r_1\right)}\mathbf{=}\left(1-\frac{g\left(r_1\right)\mathrm{dr}}{c^2}\right)$

Similarly, if we consider elevationsdata-custom-editor="chemistry" ${\mathbf{r}}_{\mathbf{1}}\mathbf{+}\mathbf{dr}$ and data-custom-editor="chemistry" ${\mathbf{r}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{dr}$, we have

$\frac{\mathbf{f}\left(r_1+2\mathrm{dr}\right)}{\mathbf{f}\left(r_1+\mathrm{dr}\right)}\mathbf{=}\left(1-\frac{g\left(r_1+\mathrm{dr}\right)\mathrm{dr}}{c^2}\right)$

We continue the process, incrementing r by dr, until we reach r₂.

$\frac{\mathbf{f}\left(r_2\right)}{\mathbf{f}\left(r_2-\mathrm{dr}\right)}\mathbf{=}\left(1-\frac{g\left(r_2-\mathrm{dr}\right)\mathrm{dr}}{c^2}\right)$

Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply $\mathbf{f}\left(r_2\right)\mathbf{/}\mathbf{f}\left(r_1\right)$. (b) Assuming that the term$\mathbf{gdr}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$ in each individual equation is very small. so that productsof such termscan be ignored, argue that the right side of the product is

$\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{\int }\mathbf{g}\left(r\right)\mathbf{dr}$

(c) Deduce$\mathbf{g}\left(r\right)$ from Newton’s universal law of gravitation, then argue that equation (2-31) follows from the result, just as (2-30) does from (2-29).

Special relativity is described as an explanation regarding how the speed of a particle affects the space, mass, and time of that particle. It applies the relativity principle in two inertial frames.

It is given that the variation of frequency in the infinitesimal interval is dr. Hence, for getting a total variation expression amongst the two positions, the small variations can be multiplied for identifying the frequency change amongst r₂ and r₁. In this particular part, the left-hand side is concerned of,

$\frac{\mathrm{f}\left({\mathrm{r}}_2\right)}{\mathrm{f}\left({\mathrm{r}}_2-\mathrm{dr}\right)}\frac{\mathrm{f}\left({\mathrm{r}}_2-\mathrm{dr}\right)}{\mathrm{f}\left({\mathrm{r}}_2-2\mathrm{dr}\right)}.......\frac{\mathrm{f}\left({\mathrm{r}}_1+2\mathrm{dr}\right)}{\mathrm{f}\left({\mathrm{r}}_1+\mathrm{dr}\right)}\frac{\mathrm{f}\left({\mathrm{r}}_1+\mathrm{dr}\right)}{\mathrm{f}\left({\mathrm{r}}_1\right)}$

Hence, except the term data-custom-editor="chemistry" $\frac{\mathrm{f}\left({\mathrm{r}}_2\right)}{\mathrm{f}\left({\mathrm{r}}_1\right)}$, all the terms are cancelled out.

Thus, the left-side of the product is simply data-custom-editor="chemistry" $\frac{\mathrm{f}\left({\mathrm{r}}_2\right)}{\mathrm{f}\left({\mathrm{r}}_1\right)}$, hence proved.

It is given that the term $\mathrm{gdr}/{\mathrm{c}}^2$’s higher term is negligible and the linear terms are taken for the expansion process.

The equation of the expansion process can be termed as:

localid="1658816155416" $$\begin{gathered} \left(1-\frac{\mathrm{g}\left({\mathrm{r}}_1\right)\mathrm{dr}}{{\mathrm{c}}^2}\right)\left(1-\frac{\mathrm{g}\left({\mathrm{r}}_1+\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}\right)\left(1-\frac{\mathrm{g}\left({\mathrm{r}}_1+2\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}\right)....\left(1-\frac{\mathrm{g}\left({\mathrm{r}}_2-2\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}\right)\left(1-\frac{\mathrm{g}\left({\mathrm{r}}_2-\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}\right) \\ =1-\left(\frac{\mathrm{g}\left({\mathrm{r}}_1\right)\mathrm{dr}}{{\mathrm{c}}^2}+\frac{\mathrm{g}\left({\mathrm{r}}_1+\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}+\frac{\mathrm{g}\left({\mathrm{r}}_1+2\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}+....+\frac{\mathrm{g}\left({\mathrm{r}}_2-2\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}+\frac{\mathrm{g}\left({\mathrm{r}}_2-\mathrm{dr}\right)\mathrm{dr}}{{\mathrm{c}}^2}\right) \\ =1-\frac{1}{{\mathrm{c}}^2}{\int }_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\mathrm{g}\left(\mathrm{r}\right)\mathrm{dr} \end{gathered}$$

Here, the integral’s basic definition has been used, that is, the small variation’s sum$\mathrm{g}\left(\mathrm{r}\right)$ in the interval r₁ and r₂.

Thus, the right-side of the product is $1-\frac{1}{{\mathrm{c}}^2}{\int }_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\mathrm{g}\left(\mathrm{r}\right)\mathrm{dr}$is proved.

The equation of the Newtonian gravity is expressed as:

$\frac{\mathrm{f}\left({\mathrm{r}}_2\right)}{\mathrm{f}\left({\mathrm{r}}_1\right)}=1-\frac{1}{{\mathrm{c}}^2}{\int }_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\mathrm{g}\left(\mathrm{r}\right)\mathrm{dr}$ ..… (1)

From the above equation, there is one value which can be obtained:

$\begin{array}{rcl}\mathrm{mg}\left(\mathrm{r}\right)& =& \frac{\mathrm{GMm}}{{\mathrm{r}}^2}\\ \mathrm{g}\left(\mathrm{r}\right)& =& \frac{\mathrm{GM}}{{\mathrm{r}}^2}\end{array}$

Substitute the above value in equation (1),

$\begin{array}{rcl}\frac{\mathrm{f}\left({\mathrm{r}}_2\right)}{\mathrm{f}\left({\mathrm{r}}_1\right)}& =& 1-\frac{1}{{\mathrm{c}}^2}{\int }_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\frac{\mathrm{GM}}{{\mathrm{r}}^2}\mathrm{dr}\\ & =& 1-\frac{\mathrm{GM}}{{\mathrm{c}}^2}{\left[-\frac{1}{\mathrm{r}}\right]}_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\\ & =& 1-\frac{\mathrm{GM}}{{\mathrm{c}}^2}\left[-\frac{1}{{\mathrm{r}}_2}-\left(-\frac{1}{{\mathrm{r}}_1}\right)\right]\\ & =& 1-\frac{\mathrm{GM}}{{\mathrm{c}}^2}\left(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right)\\ & & \\ & & \end{array}$

Equation (2-30) is expressed as:

$∆\mathrm{t}\left({\mathrm{r}}_1\right)=∆\mathrm{t}\left({\mathrm{r}}_2\right)\left(1-\frac{\mathrm{GM}}{{\mathrm{c}}^2}\left(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right)\right)$

Hence, it can be observed that equation (2-30) follows from the result.

The equation (2-31) is expressed as:

$∆\mathrm{t}\left({\mathrm{r}}_{\mathrm{Earth}}\right)=∆\mathrm{t}\left({\mathrm{r}}_{\mathrm{Satellite}}\right)\left(1-\frac{\mathrm{GM}}{{\mathrm{c}}^2}\left(\frac{1}{{\mathrm{r}}_{\mathrm{Earth}}}-\frac{1}{{\mathrm{r}}_{\mathrm{Satellite}}}\right)\right)$

Hence, it can also be observed that equation (2-31) also follows from the result.

Thus, the value ofdata-custom-editor="chemistry" $\mathrm{g}\left(\mathrm{r}\right)$ is $\frac{\mathrm{GM}}{{\mathrm{r}}^2}$.

Equation (2-31) follows from the result.