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Step-by-Step Solution
Step 1: Identify the Given Information
• Acceleration due to gravity at Earth's surface, $g = 9.8 \text{ m/s}^2$.
• We want the altitude $h$ above Earth's surface where $g$ becomes $4.9 \text{ m/s}^2$.
• Radius of the Earth, $R = 6.4 \times 10^6 \,\text{m}$.
Step 2: Recall the Formula for Acceleration due to Gravity at a Height $h$
The expression for acceleration due to gravity at a distance $(R + h)$ from the center of the Earth is given by:
$ g_h = \frac{GM}{(R + h)^2} $
where $G$ is the universal gravitational constant, and $M$ is the mass of the Earth.
Step 3: Write the Condition for the Reduced Gravity
We are told the new value of $g$ (at altitude $h$) is half of its original value at the surface:
$ g_h = \frac{g}{2} = \frac{9.8}{2} = 4.9 \,\text{m/s}^2 $
Since $g$ at Earth’s surface is $ \frac{GM}{R^2} $, we set:
$ \frac{GM}{(R+h)^2} = \frac{1}{2} \times \frac{GM}{R^2} $
Step 4: Simplify the Equation
Cancel $GM$ on both sides:
$ \frac{1}{(R+h)^2} = \frac{1}{2 R^2} $
Taking square roots:
$ R + h = \sqrt{2} \, R $
Step 5: Solve for the Altitude $h$
Rearrange for $h$:
$ h = \sqrt{2} R - R = (\sqrt{2} - 1)\,R $
Plug in $ R = 6.4 \times 10^6 \,\text{m} $:
$ h \approx (\sqrt{2} - 1) \times 6.4 \times 10^6 \,\text{m} \approx 2.6 \times 10^6 \,\text{m} $
Step 6: State the Final Answer
Hence, the altitude at which the acceleration due to gravity is half its value at the surface is approximately
$ 2.6 \times 10^6 \,\text{m} $.
