Problem Statement
Treat the Earth as a uniform sphere of mass $M = 6.0\times10^{24}\,\text{kg}$ and radius $R = 6.4\times10^{6}\,\text{m}$. Find (a) the acceleration due to gravity at the surface, (b) its value at a height $h = 5e+05\,\text{m}$ above the surface, (c) the escape velocity from the Earth’s surface, and (d) the orbital speed of a satellite at that height $h$. Take $G = 6.67\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2}$.
Given Information
- $M = 6.0\times10^{24}\,\text{kg}$ — mass of the Earth
- $R = 6.4\times10^{6}\,\text{m}$ — radius of the Earth
- $h = 5e+05\,\text{m}$ — height above the surface
- $G = 6.67\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2}$
Physical Concepts & Formulas
Newton’s law of gravitation gives the surface gravity $g=GM/R^{2}$, which falls off as the inverse square of the distance from the centre. Escape velocity is the minimum launch speed for which total mechanical energy is zero. A satellite’s orbital speed comes from equating gravitational force to the required centripetal force.
- $g = \dfrac{GM}{R^{2}}$ — surface gravity
- $g_h = \dfrac{GM}{(R+h)^{2}}$ — gravity at height $h$
- $v_e = \sqrt{\dfrac{2GM}{R}}$ — escape velocity
- $v_o = \sqrt{\dfrac{GM}{R+h}}$ — orbital speed
Step-by-Step Solution
Step 1 — Surface acceleration due to gravity: Insert Earth’s mass and radius into $g=GM/R^{2}$.
$$ g = \frac{6.67\times10^{-11}\times6.0\times10^{24}}{(6.4\times10^{6})^{2}} = 9.77\,\text{m/s}^{2} $$
Step 2 — Gravity at height $h$: Replace $R$ by $R+h$; the value decreases with altitude.
$$ g_h = \frac{GM}{(R+h)^{2}} = 8.41\,\text{m/s}^{2} $$
Step 3 — Escape velocity from the surface: Set total energy (KE + gravitational PE) to zero and solve for the launch speed.
$$ v_e = \sqrt{\frac{2GM}{R}} = 1.12e+04\,\text{m/s} $$
Step 4 — Orbital speed at height $h$: Equate gravitational force to centripetal force for a circular orbit.
$$ v_o = \sqrt{\frac{GM}{R+h}} = 7.62e+03\,\text{m/s} $$
Worked Calculation
$$ g = 9.77\,\text{m/s}^{2},\ v_e = 1.12e+04\,\text{m/s}\approx11.2\,\text{km/s},\ v_o = 7.62e+03\,\text{m/s} $$
Answer
$$ \boxed{g = 9.77\,\text{m/s}^{2},\ v_e = 1.12e+04\,\text{m/s},\ v_o = 7.62e+03\,\text{m/s}} $$
The escape velocity (~11.2 km/s) is independent of the escaping body’s mass; the orbital speed is exactly $v_e/\sqrt{2}$ for a low orbit.
Physical Interpretation
Gravity weakens with altitude, which is why high satellites move more slowly and have longer periods (Kepler’s third law). The escape speed does not depend on direction of launch, only on the depth of the gravitational well.
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