To apply the Einstein–Infeld–Hoffmann (EIH) equations to the Sun-Earth system with standard values and calculate relativistic effects, we'll use a simplified approach. We'll focus on one key relativistic effect: the perihelion precession of Earth's orbit. This effect is similar in nature to the precession observed in Mercury's orbit but is much smaller for Earth due to its larger distance from the Sun and smaller orbital eccentricity.

### Key Parameters

1. **Masses:**

- \( M_{\text{Sun}} \approx 1.989 \times 10^{30} \) kg

- \( M_{\text{Earth}} \approx 5.972 \times 10^{24} \) kg

2. **Orbital Characteristics:**

- Semi-major axis of Earth's orbit \( a \approx 1.496 \times 10^{11} \) meters (1 AU)

- Orbital eccentricity \( e \approx 0.0167 \)

- Orbital period \( T \approx 365.25 \) days \( = 3.156 \times 10^7 \) seconds

3. **Constants:**

- Gravitational constant \( G \approx 6.674 \times 10^{-11} \) m³ kg⁻¹ s⁻²

- Speed of light \( c \approx 2.998 \times 10^8 \) m/s

### Perihelion Precession Calculation

In general relativity, the perihelion precession of a planet’s orbit can be approximated by:

\[

\Delta \omega = \frac{6 \pi G M_{\text{Sun}}}{c^2 a (1 - e^2)}

\]

where:

- \(\Delta \omega\) is the precession per orbit,

- \(a\) is the semi-major axis,

- \(e\) is the orbital eccentricity.

Substitute the standard values into the formula:

\[

\Delta \omega = \frac{6 \pi \times 6.674 \times 10^{-11} \times 1.989 \times 10^{30}}{(2.998 \times 10^8)^2 \times 1.496 \times 10^{11} \times (1 - 0.0167^2)}

\]

### Performing the Calculation

1. **Calculate the denominator:**

\[

c^2 = (2.998 \times 10^8)^2 \approx 8.988 \times 10^{16} \text{ m}^2/\text{s}^2

\]

\[

c^2 \times a \times (1 - e^2) = 8.988 \times 10^{16} \times 1.496 \times 10^{11} \times (1 - 0.00027889) \approx 1.349 \times 10^{28}

\]

2. **Calculate the numerator:**

\[

6 \pi G M_{\text{Sun}} \approx 6 \times 3.142 \times 6.674 \times 10^{-11} \times 1.989 \times 10^{30} \approx 1.577 \times 10^{20}

\]

3. **Divide the numerator by the denominator:**

\[

\Delta \omega = \frac{1.577 \times 10^{20}}{1.349 \times 10^{28}} \approx 1.17 \times 10^{-8} \text{ radians per orbit}

\]

4. **Convert to degrees:**

\[

\Delta \omega \approx 1.17 \times 10^{-8} \text{ radians} \times \frac{180}{\pi} \approx 6.7 \times 10^{-7} \text{ degrees per orbit}

\]

### Summary

The perihelion precession of Earth's orbit due to relativistic effects is approximately \(6.7 \times 10^{-7}\) degrees per orbit. This value is exceedingly small compared to the observable effects in the inner solar system (such as Mercury's perihelion precession) but demonstrates how relativistic corrections are applied in practice.