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The acceleration due to gravity at a point 'X' at a height of 100 m from the surface of earth is 10 m/s2. Imagine the earth suddenly shrinks uniformly to half of its present size without losing any mass. What will be the value of acceleration due to gravity at the point 'X'?
- a)2.5 m/s2
- b)5 m/s2
- c)10 m/s2
- d)Zero
Correct answer is option 'C'. Can you explain this answer?
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The acceleration due to gravity at a point X at a height of 100 m from...
Explanation:
When the earth shrinks uniformly to half of its present size without losing any mass, the distance from the point X to the center of the earth will also be halved.
Let's assume the original radius of the earth is R.
After shrinking, the radius of the earth becomes R/2.
Step 1: Calculating the original acceleration due to gravity at point X
The acceleration due to gravity at a point above the surface of the earth can be calculated using the formula:
g = (G * M) / (R + h)^2
where:
g is the acceleration due to gravity,
G is the gravitational constant,
M is the mass of the earth,
R is the original radius of the earth, and
h is the height from the surface of the earth.
Given that the acceleration due to gravity at point X is 10 m/s^2 and the height from the surface of the earth is 100 m, we can substitute these values into the formula:
10 = (G * M) / (R + 100)^2
Step 2: Calculating the new acceleration due to gravity at point X
After the shrinkage, the new radius of the earth is R/2.
Using the same formula, we can now calculate the new acceleration due to gravity at point X:
g' = (G * M) / (R/2 + 100)^2
Simplifying the equation:
g' = (G * M) / ((R/2 + 100)^2)
Since the mass of the earth remains the same, we can cancel it out:
g' = (G * M) / ((R/2 + 100)^2) = (G * M) / (R^2/4 + 200R + 10000)
Step 3: Comparing the original and new acceleration due to gravity
To compare the original and new acceleration due to gravity, we divide the new acceleration by the original acceleration:
g'/g = [(G * M) / (R^2/4 + 200R + 10000)] / [(G * M) / (R + 100)^2]
Canceling out the mass and gravitational constant:
g'/g = (R + 100)^2 / (R^2/4 + 200R + 10000)
Expanding and simplifying the equation:
g'/g = (R^2 + 200R + 10000) / (R^2/4 + 200R + 10000)
Since the original radius of the earth is R, we can substitute R^2 with R^2/4:
g'/g = (R^2/4 + 200R + 10000) / (R^2/4 + 200R + 10000)
Canceling out the common terms:
g'/g = 1
Therefore, the new acceleration due to gravity at point X is equal to the original acceleration due to gravity, which is 10 m/s^2.
Hence, the correct answer is option C) 10 m/s^2.
When the earth shrinks uniformly to half of its present size without losing any mass, the distance from the point X to the center of the earth will also be halved.
Let's assume the original radius of the earth is R.
After shrinking, the radius of the earth becomes R/2.
Step 1: Calculating the original acceleration due to gravity at point X
The acceleration due to gravity at a point above the surface of the earth can be calculated using the formula:
g = (G * M) / (R + h)^2
where:
g is the acceleration due to gravity,
G is the gravitational constant,
M is the mass of the earth,
R is the original radius of the earth, and
h is the height from the surface of the earth.
Given that the acceleration due to gravity at point X is 10 m/s^2 and the height from the surface of the earth is 100 m, we can substitute these values into the formula:
10 = (G * M) / (R + 100)^2
Step 2: Calculating the new acceleration due to gravity at point X
After the shrinkage, the new radius of the earth is R/2.
Using the same formula, we can now calculate the new acceleration due to gravity at point X:
g' = (G * M) / (R/2 + 100)^2
Simplifying the equation:
g' = (G * M) / ((R/2 + 100)^2)
Since the mass of the earth remains the same, we can cancel it out:
g' = (G * M) / ((R/2 + 100)^2) = (G * M) / (R^2/4 + 200R + 10000)
Step 3: Comparing the original and new acceleration due to gravity
To compare the original and new acceleration due to gravity, we divide the new acceleration by the original acceleration:
g'/g = [(G * M) / (R^2/4 + 200R + 10000)] / [(G * M) / (R + 100)^2]
Canceling out the mass and gravitational constant:
g'/g = (R + 100)^2 / (R^2/4 + 200R + 10000)
Expanding and simplifying the equation:
g'/g = (R^2 + 200R + 10000) / (R^2/4 + 200R + 10000)
Since the original radius of the earth is R, we can substitute R^2 with R^2/4:
g'/g = (R^2/4 + 200R + 10000) / (R^2/4 + 200R + 10000)
Canceling out the common terms:
g'/g = 1
Therefore, the new acceleration due to gravity at point X is equal to the original acceleration due to gravity, which is 10 m/s^2.
Hence, the correct answer is option C) 10 m/s^2.
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The acceleration due to gravity at a point X at a height of 100 m from the surface of earth is 10 m/s2. Imagine the earth suddenly shrinks uniformly to half of its present size without losing any mass. What will be the value of acceleration due to gravity at the point X?a)2.5 m/s2b)5 m/s2c)10 m/s2d)ZeroCorrect answer is option 'C'. Can you explain this answer?
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