Calculate the work required to lift a mass m

Calculate the work required to lift a mass $m$ from the Earth's surface to height $h$, taking into account that gravitational force decreases with distance from the Earth's center. What is the work when $h\to\infty$?

Step 1: Work integral

\[W=\int_{R_E}^{R_E+h}G\frac{Mm}{r^2}\,dr=GMm\left[-\frac1r\right]_{R_E}^{R_E+h}=GMm\left(\frac{1}{R_E}-\frac{1}{R_E+h}\right).\]

Step 2: Simplify

\[W=GMm\cdot\frac{h}{R_E(R_E+h)}.\]

\[\boxed{W=\frac{GMmh}{R_E(R_E+h)}}\]

Step 3: Rewrite using surface gravity

With $g_0=GM/R_E^2$:

\[\boxed{W=\frac{mg_0R_Eh}{R_E+h}}\]

Step 4: Small-height check

For $h\ll R_E$: $W\approx\dfrac{mg_0R_Eh}{R_E}=mg_0h$, the familiar constant-gravity formula.

Step 5: The limit $h\to\infty$

\[W_\infty=\lim_{h\to\infty}\frac{mg_0R_Eh}{R_E+h}=mg_0R_E=\frac{GMm}{R_E},\]

a finite value even though the distance is infinite, because gravity weakens as $1/r^2$. This matches direct evaluation of $\int_{R_E}^{\infty}GMm/r^2\,dr$.

\[\boxed{W_\infty=mg_0R_E}\]

Step 6: Escape velocity check

Setting $\tfrac12mv_e^2=W_\infty$ gives $v_e=\sqrt{2g_0R_E}$, which for Earth's values evaluates to about $11.2\ \text{km/s}$ — the known escape velocity, confirming the result.