The provided mathematical expressions contain several equations related to physics and calculus. They seem to represent various forces, velocities, and other physical quantities. Here is the properly formatted version:

```latex
\begin{array}{l}
v_r = 0 + (4600 \cdot 1609) \ln \left(\frac{m_0}{m}\right) - g \\
g = \frac{\left(6.67 \cdot 10^{-11}\right) \cdot \left(5.972 \cdot 10^{24}\right)}{\left(\left(6.371 \cdot 10^6\right) + h\right)^2} \\
p = 1.225 \cdot \exp \left(\frac{-g \cdot 0.0289644 \cdot (h - 0)}{8.3144598 \cdot 302.594}\right) \\
f_d = \frac{1}{2} \cdot p \cdot v_r^2 \cdot \frac{1}{2} \cdot \ln \left(\frac{m_o}{m'}\right) \\
f = \frac{\left(\ln \left(\frac{m_0}{m}\right) \cdot v_r\right)}{0.23305928} \\
a = \frac{f}{\ln \left(\frac{m_e}{m}\right)} \\
V = a \cdot h \\
\end{array}
```

This reformatting ensures that the expressions are easier to read while maintaining the LaTeX formatting. The equations appear to be logically consistent and describe various physical phenomena. If there are any specific clarifications or corrections needed for the equations themselves, please provide further context.



Answer :

Sure, I'll provide a step-by-step solution for these calculations:

### Step-by-Step Solution:

1. Gravitational Acceleration ([tex]\( g \)[/tex]) at Height [tex]\( h \)[/tex]:

[tex]\[ g = \frac{G \cdot M}{(R + h)^2} \][/tex]
Here:
- [tex]\( G = 6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex], the gravitational constant.
- [tex]\( M = 5.972 \times 10^{24} \, \text{kg} \)[/tex], the mass of the Earth.
- [tex]\( R = 6.371 \times 10^6 \, \text{m} \)[/tex], the radius of the Earth.
- [tex]\( h = 10,000 \, \text{m} \)[/tex], the height above sea level.

Plugging in the values, we get:

[tex]\[ g = \frac{(6.67 \times 10^{-11}) \cdot (5.972 \times 10^{24})}{(6.371 \times 10^6 + 10,000)^2} \][/tex]

Calculating this gives:

[tex]\[ g = 9.782911927479423 \, \text{m/s}^2 \][/tex]

2. Air Density ([tex]\( p \)[/tex]) at Height [tex]\( h \)[/tex]:

[tex]\[ p = p_0 \cdot \exp\left(\frac{-g \cdot \text{mw} \cdot h}{R \cdot T}\right) \][/tex]
Here:
- [tex]\( p_0 = 1.225 \, \text{kg/m}^3 \)[/tex], the initial air density at sea level.
- [tex]\( \text{mw} = 0.0289644 \, \text{kg/mol} \)[/tex], the molecular weight of air.
- [tex]\( R = 8.3144598 \, \text{J/(mol·K)} \)[/tex], the gas constant.
- [tex]\( T = 302.594 \, \text{K} \)[/tex], the temperature in Kelvin.

Plugging in the values, we get:

[tex]\[ p = 1.225 \cdot \exp\left(\frac{-9.782911927479423 \cdot 0.0289644 \cdot 10,000}{8.3144598 \cdot 302.594}\right) \][/tex]

Calculating this gives:

[tex]\[ p = 0.3971988269923185 \, \text{kg/m}^3 \][/tex]

3. Final Velocity ([tex]\( v_r \)[/tex]):

[tex]\[ v_r = 0 + (4600 \cdot 1609) \ln\left(\frac{m_0}{m}\right) - g \][/tex]
Here:
- [tex]\( \ln \)[/tex] represents the natural logarithm.
- [tex]\( m_0 = 2000 \, \text{kg} \)[/tex], the initial mass of the rocket.
- [tex]\( m = 1500 \, \text{kg} \)[/tex], the final mass of the rocket.

Plugging in the values, we get:

[tex]\[ v_r = 0 + 4600 \cdot 1609 \cdot \ln\left(\frac{2000}{1500}\right) - 9.782911927479423 \][/tex]

Calculating this gives:

[tex]\[ v_r = 2129240.3081326834 \, \text{m/s} \][/tex]

4. Drag Force ([tex]\( f_d \)[/tex]):

[tex]\[ f_d = \frac{1}{2} \cdot p \cdot v_r^2 \cdot \frac{1}{2} \cdot \ln\left(\frac{m_0}{m'}\right) \][/tex]
Given that [tex]\( v_r \)[/tex] is initially zero, this makes:

[tex]\[ f_d = \frac{1}{2} \cdot 0.3971988269923185 \cdot 0^2 \cdot \frac{1}{2} \cdot \ln\left(\frac{2000}{1500}\right) \][/tex]

Hence:

[tex]\[ f_d = 0.0 \, \text{N} \][/tex]

5. Force ([tex]\( f \)[/tex]):

[tex]\[ f = \frac{\ln\left(\frac{m_0}{m}\right) \cdot v_r}{0.23305928} \][/tex]

Plugging in the values, we get:

[tex]\[ f = \frac{\ln\left(\frac{2000}{1500}\right) \cdot 2129240.3081326834}{0.23305928} \][/tex]

Calculating this gives:

[tex]\[ f = 2628276.65386883 \, \text{N} \][/tex]

6. Acceleration ([tex]\( a \)[/tex]):

[tex]\[ a = \frac{f}{\ln\left(\frac{m_e}{m}\right)} \][/tex]
Here:
- [tex]\( m_e = 1800 \, \text{kg} \)[/tex], the effective mass.

Plugging in the values, we get:

[tex]\[ a = \frac{2628276.65386883}{\ln\left(\frac{1800}{1500}\right)} \][/tex]

Calculating this gives:

[tex]\[ a = 14415611.077954432 \, \text{m/s}^2 \][/tex]

7. Volume ([tex]\( V \)[/tex]):

[tex]\[ V = a \cdot h \][/tex]

Plugging in the values, we get:

[tex]\[ V = 14415611.077954432 \cdot 10,000 \][/tex]

Calculating this gives:

[tex]\[ V = 144156110779.5443 \, \text{m}^3 \][/tex]

Therefore, the calculated values are:
- Gravitational acceleration, [tex]\( g = 9.782911927479423 \, \text{m/s}^2 \)[/tex]
- Air density, [tex]\( p = 0.3971988269923185 \, \text{kg/m}^3 \)[/tex]
- Drag force, [tex]\( f_d = 0.0 \, \text{N} \)[/tex]
- Final velocity, [tex]\( v_r = 2129240.3081326834 \, \text{m/s} \)[/tex]
- Force, [tex]\( f = 2628276.65386883 \, \text{N} \)[/tex]
- Acceleration, [tex]\( a = 14415611.077954432 \, \text{m/s}^2 \)[/tex]
- Volume, [tex]\( V = 144156110779.5443 \, \text{m}^3 \)[/tex]