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Sagot :
Sure, let's analyze the correctness of the formula [tex]\(\sqrt{\frac{R}{2 G M}}\)[/tex] for escape velocity using dimensional analysis.
### Step-by-Step Dimensional Analysis:
1. Understand the Dimensions of Each Quantity:
- [tex]\( R \)[/tex] (radius) has the dimension of length [tex]\([L]\)[/tex].
- [tex]\( G \)[/tex] (gravitational constant) has the dimensions [tex]\([L^3 M^{-1} T^{-2}]\)[/tex].
- [tex]\( M \)[/tex] (mass) has the dimension [tex]\([M]\)[/tex].
2. Combine the Dimensions in the Given Formula:
The formula provided is [tex]\(\frac{R}{2 G M}\)[/tex]. We need to analyze the dimensions of this expression inside the square root:
[tex]\[ \frac{R}{2 G M} \][/tex]
3. Analyze the Individual Terms in the Denominator:
- The constant 2 is dimensionless.
- The dimension of [tex]\( G \)[/tex] is [tex]\([L^3 M^{-1} T^{-2}]\)[/tex].
- The dimension of [tex]\( M \)[/tex] is [tex]\([M]\)[/tex].
4. Combine the Dimensions in the Denominator:
Combine the dimensions of [tex]\( G \)[/tex] and [tex]\( M \)[/tex]:
[tex]\[ 2 \times G \times M = 2 \times [L^3 M^{-1} T^{-2}] \times [M] = [L^3 M^{-1} T^{-2}] \times [M] = [L^3 T^{-2}] \][/tex]
5. Evaluate the Dimensions of the Entire Expression:
Next, we evaluate the dimensions of the expression [tex]\( \frac{R}{2 G M} \)[/tex]:
[tex]\[ \frac{R}{2 G M} = \frac{[L]}{[L^3 T^{-2}]} = [L] \times [L^{-3} T^{2}] = [L^{-2} T^{2}] \][/tex]
6. Take the Square Root of the Dimensions:
Finally, take the square root of the dimensions of the expression:
[tex]\[ \sqrt{[L^{-2} T^2]} = [L^{-1} T] \][/tex]
### Conclusion:
The dimension of the expression [tex]\(\sqrt{\frac{R}{2 G M}}\)[/tex] is found to be [tex]\([L^{-1} T]\)[/tex].
For escape velocity, the correct dimension should be that of velocity, which is [tex]\([L T^{-1}]\)[/tex]. The dimension [tex]\([L^{-1} T]\)[/tex] does not match this, indicating that the formula [tex]\(\sqrt{\frac{R}{2 G M}}\)[/tex] is dimensionally incorrect for escape velocity.
Thus, the student’s formula [tex]\(\sqrt{\frac{R}{2 G M}}\)[/tex] is not correct for calculating escape velocity.
### Step-by-Step Dimensional Analysis:
1. Understand the Dimensions of Each Quantity:
- [tex]\( R \)[/tex] (radius) has the dimension of length [tex]\([L]\)[/tex].
- [tex]\( G \)[/tex] (gravitational constant) has the dimensions [tex]\([L^3 M^{-1} T^{-2}]\)[/tex].
- [tex]\( M \)[/tex] (mass) has the dimension [tex]\([M]\)[/tex].
2. Combine the Dimensions in the Given Formula:
The formula provided is [tex]\(\frac{R}{2 G M}\)[/tex]. We need to analyze the dimensions of this expression inside the square root:
[tex]\[ \frac{R}{2 G M} \][/tex]
3. Analyze the Individual Terms in the Denominator:
- The constant 2 is dimensionless.
- The dimension of [tex]\( G \)[/tex] is [tex]\([L^3 M^{-1} T^{-2}]\)[/tex].
- The dimension of [tex]\( M \)[/tex] is [tex]\([M]\)[/tex].
4. Combine the Dimensions in the Denominator:
Combine the dimensions of [tex]\( G \)[/tex] and [tex]\( M \)[/tex]:
[tex]\[ 2 \times G \times M = 2 \times [L^3 M^{-1} T^{-2}] \times [M] = [L^3 M^{-1} T^{-2}] \times [M] = [L^3 T^{-2}] \][/tex]
5. Evaluate the Dimensions of the Entire Expression:
Next, we evaluate the dimensions of the expression [tex]\( \frac{R}{2 G M} \)[/tex]:
[tex]\[ \frac{R}{2 G M} = \frac{[L]}{[L^3 T^{-2}]} = [L] \times [L^{-3} T^{2}] = [L^{-2} T^{2}] \][/tex]
6. Take the Square Root of the Dimensions:
Finally, take the square root of the dimensions of the expression:
[tex]\[ \sqrt{[L^{-2} T^2]} = [L^{-1} T] \][/tex]
### Conclusion:
The dimension of the expression [tex]\(\sqrt{\frac{R}{2 G M}}\)[/tex] is found to be [tex]\([L^{-1} T]\)[/tex].
For escape velocity, the correct dimension should be that of velocity, which is [tex]\([L T^{-1}]\)[/tex]. The dimension [tex]\([L^{-1} T]\)[/tex] does not match this, indicating that the formula [tex]\(\sqrt{\frac{R}{2 G M}}\)[/tex] is dimensionally incorrect for escape velocity.
Thus, the student’s formula [tex]\(\sqrt{\frac{R}{2 G M}}\)[/tex] is not correct for calculating escape velocity.
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