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| Acceleration due to Gravity on Moon |
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| The expression for acceleration due to gravity is |
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| Where G is the universal gravitation constant, M is the mass of the heavenly body which produces acceleration in a body and R is the radius of the heavenly body. |
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| The equation for g shows that the value of acceleration due to gravity depends on the mass and radius of the heavenly body and hence will be different for different heavenly bodies. |
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| Let us now derive a relation between the acceleration due to gravity on moon (gm) and acceleration due to gravity on Earth (ge). |
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| Where Me and Re are the mass and radius of the Earth respectively. |
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| Where Mm and Rm are the mass and radius of the moon respectively. |
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| Divide eq (1) by eq (2) |
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| We know that mass of the Earth is 100 times that of the moon and its radius is four times that of the moon. |
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| i.e. |
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| Me = 100 Mm |
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| Re = 4 Rm. |
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| Which means that acceleration due to gravity on moon is 1/6th that on the Earth. |
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 Freely falling object: Is the object which moves towards the Earth due to force of gravity. |
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Acceleration due to gravity: It is the acceleration produced in an object due to force of gravity. |
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Equations of motion for a freely falling object are: |
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Equations of motion for an object moving against gravity are: |
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Acceleration due to gravity varies from place to place. |
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Acceleration due to gravity is zero at the centre of the Earth. |
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Acceleration due to gravity is independent of the mass of the object. |
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Newton's third law of motion is applicable to force of gravitation. |
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| 1. The Earth's gravitational force causes an acceleration of 5 m/s2 in a 1kg mass somewhere in space. How much will the acceleration of a 3kg mass be at the same place? |
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| Solution: |
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| The acceleration produced in any body due to the gravitational pull of the Earth does not depend on the mass of the body. So the acceleration produced in the 3kg mass will also be 5 m/s2. |
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| 2. Calculate the height of a bridge if a stone dropped from it takes six seconds to touch the surface of water. |
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| Solution: |
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| Initial velocity (u) = 0 |
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| Time taken (t) = 6 seconds |
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| Acceleration due to gravity (g) = 9.8 m/s2 |
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| We make use of second equation of motion |
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| 3. A stone projected vertically upward, takes 5 seconds to reach the highest point. What is the initial velocity of the stone? |
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| Solution: |
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| Time taken = 5 seconds |
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| Final velocity (v)=0 (at maximum height velocity will be equal to zero) |
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| Acceleration due to gravity (g) = -9.8 m/s2 |
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| Initial velocity = 49 m/s. |
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