step:

**8×π×6.67430×1.0×**

**10**

−11

**×**

**10**

−26

**3**




**≈1.761×**

**10**

−36

8.3×

10

−3

3




≈2.767×

10

−3

Finally, sum these two values to get the Friedmann equation for a flat universe.







Hubble constant 




To find the Hubble Parameter 

H

, you need to take the square root of the value given:

H=

2.767×

10

−3

−

−

−

−

−

−

−

−

−

−

√

Calculating the square root of 

2.767×

10

−3

:

H≈

2.767

−

−

−

−

√

×

10

−3

−

−

−

−

√

H≈1.664×

10

−1

s

−1

Therefore, the Hubble Parameter 

H

is approximately 

0.1664

s

−1

.







To rearrange the Friedmann equation to solve for 

Λ

, follow these steps: Given the Friedmann equation:

Λ

3




=

H

2

−

8πGρ

3




To solve for 

Λ

, multiply both sides by 3:

Λ=3(

H

2

−

8πGρ

3




)

Substitute the values for 

H

2

 and 

ρ

into the equation to find 

Λ

:

Λ

=3

⎛

⎝

⎜

⎜

⎜

(1.664×

10

−1

)

2

−

8×π×6.67430×

10

−11

×1.0

×

10

−26

3




⎞

⎠

⎟

⎟

⎟

Calculate the value of 

Λ

 using the given values.

To solve for the cosmological constant \(\Lambda\) using the Friedmann equation for a flat universe, we follow the steps given and perform the necessary calculations. The relevant equations and steps are as follows:

### Step-by-Step Calculation

1. **Given Values:**
   - \(G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)
   - \(\rho = 1.0 \times 10^{-26} \, \text{kg/m}^3\)
   - \(H = 0.1664 \, \text{s}^{-1}\) (calculated from the given steps)
   - \(\pi \approx 3.14159\)

2. **Friedmann Equation for a Flat Universe:**
   \[
   \Lambda = 3 \left( H^2 - \frac{8 \pi G \rho}{3} \right)
   \]

3. **Calculate \(H^2\):**
   \[
   H^2 = (0.1664 \, \text{s}^{-1})^2 = 0.0277 \, \text{s}^{-2}
   \]

4. **Calculate the term \(\frac{8 \pi G \rho}{3}\):**
   \[
   \frac{8 \pi G \rho}{3} = \frac{8 \times 3.14159 \times 6.67430 \times 10^{-11} \times 1.0 \times 10^{-26}}{3}
   \]
   \[
   \frac{8 \pi G \rho}{3} = \frac{8 \times 3.14159 \times 6.67430 \times 10^{-37}}{3}
   \]
   \[
   \frac{8 \pi G \rho}{3} \approx \frac{167.5518 \times 10^{-37}}{3} \approx 55.8506 \times 10^{-37} \approx 5.58506 \times 10^{-36} \, \text{s}^{-2}
   \]

5. **Subtract the terms and multiply by 3 to find \(\Lambda\):**
   \[
   \Lambda = 3 \left( 0.0277 \, \text{s}^{-2} - 5.58506 \times 10^{-36} \, \text{s}^{-2} \right)
   \]
   \[
   \Lambda = 3 \left( 0.0277 - 0.00000000000000000000000558506 \right)
   \]
   Since \(5.58506 \times 10^{-36}\) is extremely small compared to 0.0277, it can be approximated:
   \[
   \Lambda \approx 3 \times 0.0277 \approx 0.0831 \, \text{s}^{-2}
   \]

### Final Answer

The cosmological constant \(\Lambda\) is approximately \(0.0831 \, \text{s}^{-2}\).

### Questions for Further Exploration

1. How does the cosmological constant \(\Lambda\) affect the expansion rate of the universe?
2. What observational evidence supports the current value of the Hubble constant \(H\)?
3. How does dark energy relate to the cosmological constant in modern cosmology?
4. What are the implications of a non-zero cosmological constant for the fate of the universe?
5. How do different cosmological models adjust the value of \(\Lambda\) and \(H\)?
6. Can variations in \(\rho\) (matter density) significantly alter the value of \(\Lambda\)?
7. What role does the cosmological constant play in the context of general relativity?
8. How can future measurements improve the accuracy of \(\Lambda\) and \(H\)?

### Tip

When dealing with extremely small or large values in cosmology, always consider the significance and scale of these values in relation to the problem at hand to avoid unnecessary complexity in your calculations.

Learn how to calculate the cosmological constant \(\Lambda\) and Hubble parameter \(H\) using the Friedmann equation for a flat universe. This problem involves complex calculations in cosmology, including detailed steps for deriving \(\Lambda\) and determining \(H\). Suitable for advanced physics enthusiasts and students.

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