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Step-by-Step Solution
Step 1: Identify the Relevant Physical Concept
We use the concept of gravitational potential energy (GPE) for a mass placed at a distance $r$ from the center of the Earth. The formula for GPE (with the reference taken at infinity) is:
$U = -\dfrac{GMm}{r}$
where:
$G$ is the universal gravitational constant,
$M$ is the mass of the Earth,
$m$ is the mass of the body,
$r$ is the distance from the center of the Earth.
Step 2: Write Down GPE at the Earth’s Surface
When the body is on the Earth’s surface, $r = R$ (Earth’s radius). Thus, its gravitational potential energy is:
$U_{\text{surface}} = -\dfrac{GMm}{R}$
Step 3: Write Down GPE at the Height $h = 3R$
The new distance from the center of the Earth is $R + h = R + 3R = 4R$. Hence, the gravitational potential energy at this height is:
$U_{\text{height}} = -\dfrac{GMm}{4R}$
Step 4: Calculate the Change in Potential Energy
The change in gravitational potential energy ($\Delta U$) is given by:
$\Delta U = U_{\text{height}} - U_{\text{surface}}$
Substitute the expressions from above:
$\Delta U = \left(-\dfrac{GMm}{4R}\right) \;-\; \left(-\dfrac{GMm}{R}\right)$
Simplify term by term:
$\Delta U = -\dfrac{GMm}{4R} + \dfrac{GMm}{R}$
$\Delta U = \left(\dfrac{4GMm}{4R} - \dfrac{GMm}{4R}\right)$
$\Delta U = \dfrac{3GMm}{4R}$
Step 5: Express in Terms of $mgR$
We know that $GM = gR^2$ (approximately, for Earth). Hence,
$\dfrac{GMm}{R} = mgR$
Therefore,
$\Delta U = \dfrac{3}{4} \cdot mgR$
So, the change in gravitational potential energy is:
$\boxed{\dfrac{3}{4} \, mgR}$
