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G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.?
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G at a depth d reduces to 25% from the value at the surface . Calculat...
**Solution:**
To solve this problem, we can use the concept of the gravitational field strength at different depths within the Earth.
Let's assume that the gravitational field strength at the Earth's surface is denoted by g₀, and the gravitational field strength at a depth d is denoted by g.
According to the problem, the gravitational field strength at a depth d reduces to 25% (or 0.25 times) of the value at the surface. Mathematically, we can write this as:
g = 0.25 * g₀
Now, we need to find the depth d at which this reduction occurs. We can use the formula for the gravitational field strength inside a uniform sphere:
g = (4/3) * π * G * ρ * (R - d)
Where:
- g is the gravitational field strength at depth d,
- G is the gravitational constant (approximately 6.67430 × 10^(-11) N m²/kg²),
- ρ is the average density of the Earth,
- R is the radius of the Earth.
We can substitute the given value of g (0.25 * g₀) into the formula and solve for d.
**Calculating the depth d:**
Let's substitute the values into the formula and simplify:
0.25 * g₀ = (4/3) * π * G * ρ * (R - d)
Divide both sides of the equation by (4/3) * π * G * ρ:
(0.25 * g₀) / ((4/3) * π * G * ρ) = R - d
Rearrange the equation:
d = R - (0.25 * g₀) / ((4/3) * π * G * ρ)
Now, we can substitute the known values into the equation:
- R = 6400 km
- g₀ = 9.8 m/s² (approximate value for the gravitational field strength at the Earth's surface)
- G = 6.67430 × 10^(-11) N m²/kg² (gravitational constant)
We need to convert the radius of the Earth to meters and the depth to kilometers to maintain consistent units.
**Substituting the values:**
- R = 6400 km = 6400 * 1000 m = 6,400,000 m
- g₀ = 9.8 m/s²
- G = 6.67430 × 10^(-11) N m²/kg²
Now, let's calculate the value of d:
d = 6,400,000 m - (0.25 * 9.8 m/s²) / ((4/3) * π * (6.67430 × 10^(-11) N m²/kg²) * ρ)
The average density of the Earth, ρ, is approximately 5,515 kg/m³.
**Calculating the final value of d:**
Substituting the values and calculating:
d ≈ 6,400,000 m - (0.25 * 9.8 m/s²) / ((4/3) * π * (6.67430 × 10^(-11) N m²/kg²) * 5,515 kg/m³)
After performing the calculations, we find:
To solve this problem, we can use the concept of the gravitational field strength at different depths within the Earth.
Let's assume that the gravitational field strength at the Earth's surface is denoted by g₀, and the gravitational field strength at a depth d is denoted by g.
According to the problem, the gravitational field strength at a depth d reduces to 25% (or 0.25 times) of the value at the surface. Mathematically, we can write this as:
g = 0.25 * g₀
Now, we need to find the depth d at which this reduction occurs. We can use the formula for the gravitational field strength inside a uniform sphere:
g = (4/3) * π * G * ρ * (R - d)
Where:
- g is the gravitational field strength at depth d,
- G is the gravitational constant (approximately 6.67430 × 10^(-11) N m²/kg²),
- ρ is the average density of the Earth,
- R is the radius of the Earth.
We can substitute the given value of g (0.25 * g₀) into the formula and solve for d.
**Calculating the depth d:**
Let's substitute the values into the formula and simplify:
0.25 * g₀ = (4/3) * π * G * ρ * (R - d)
Divide both sides of the equation by (4/3) * π * G * ρ:
(0.25 * g₀) / ((4/3) * π * G * ρ) = R - d
Rearrange the equation:
d = R - (0.25 * g₀) / ((4/3) * π * G * ρ)
Now, we can substitute the known values into the equation:
- R = 6400 km
- g₀ = 9.8 m/s² (approximate value for the gravitational field strength at the Earth's surface)
- G = 6.67430 × 10^(-11) N m²/kg² (gravitational constant)
We need to convert the radius of the Earth to meters and the depth to kilometers to maintain consistent units.
**Substituting the values:**
- R = 6400 km = 6400 * 1000 m = 6,400,000 m
- g₀ = 9.8 m/s²
- G = 6.67430 × 10^(-11) N m²/kg²
Now, let's calculate the value of d:
d = 6,400,000 m - (0.25 * 9.8 m/s²) / ((4/3) * π * (6.67430 × 10^(-11) N m²/kg²) * ρ)
The average density of the Earth, ρ, is approximately 5,515 kg/m³.
**Calculating the final value of d:**
Substituting the values and calculating:
d ≈ 6,400,000 m - (0.25 * 9.8 m/s²) / ((4/3) * π * (6.67430 × 10^(-11) N m²/kg²) * 5,515 kg/m³)
After performing the calculations, we find:
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G at a depth d reduces to 25% from the value at the surface . Calculat...
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G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.?
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G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.?.
G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for G at a depth d reduces to 25% from the value at the surface . Calculate d if radius of Earth is 6400 km.?.
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