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If R is the radius of the Earth then the...

If R is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by `20%` is

A

`((sqrt5)/(2)-1)R`

B

`((sqrt5)/(2)+1)R`

C

`(5sqrt2-1)R`

D

`(5sqrt2+1)R`

Text Solution

AI Generated Solution

The correct Answer is:
To find the height above the Earth's surface at which the acceleration due to gravity decreases by 20%, we can follow these steps: ### Step 1: Understand the relationship between gravity at the surface and at height The acceleration due to gravity at the surface of the Earth is given by: \[ g = \frac{GM}{R^2} \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth. ### Step 2: Define the gravity at height \( h \) At a height \( h \) above the Earth's surface, the acceleration due to gravity \( g' \) is given by: \[ g' = \frac{GM}{(R + h)^2} \] ### Step 3: Set up the equation for the decrease in gravity According to the problem, the acceleration due to gravity decreases by 20%. Therefore, we can express this as: \[ g' = g - 0.2g = 0.8g \] ### Step 4: Substitute the expressions for \( g \) and \( g' \) We can substitute the expressions for \( g \) and \( g' \) into the equation: \[ \frac{GM}{(R + h)^2} = 0.8 \cdot \frac{GM}{R^2} \] ### Step 5: Cancel \( GM \) from both sides Since \( GM \) is common on both sides, we can cancel it out: \[ \frac{1}{(R + h)^2} = 0.8 \cdot \frac{1}{R^2} \] ### Step 6: Rearrange the equation Rearranging gives us: \[ (R + h)^2 = \frac{R^2}{0.8} \] ### Step 7: Simplify the equation Taking the square root of both sides: \[ R + h = \frac{R}{\sqrt{0.8}} \] ### Step 8: Solve for \( h \) Now, isolate \( h \): \[ h = \frac{R}{\sqrt{0.8}} - R \] \[ h = R \left( \frac{1}{\sqrt{0.8}} - 1 \right) \] ### Step 9: Simplify further We can simplify \( \sqrt{0.8} \): \[ \sqrt{0.8} = \frac{2}{\sqrt{5}} \] Thus, \[ h = R \left( \frac{\sqrt{5}}{2} - 1 \right) \] ### Final Result The height \( h \) above the Earth's surface at which the acceleration due to gravity decreases by 20% is: \[ h = R \left( \frac{\sqrt{5}}{2} - 1 \right) \] ---
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