Problem Statement
Solve the gravitation problem: A satellite orbits Earth at radius $r = 7 \times 10^6$ m. Find the orbital speed. ($g = 9.8$ m/s² at Earth’s surface, $R_E = 6.4 \times 10^6$ m) $v = \sqrt{GM_E/r}$; use $GM_E = gR_E^2$ Step 1: $GM_E = gR_E^2 = 9.8 \times (6.4\times10^6)^2 = 9.8 \times 4.096\times10^{13} = 4.01\times10^{14}$ m³/s².
Given Information
- Masses $M$ (planet/star) and $m$ (object/satellite)
- Orbital radius $r$ or altitude $h$
- $G = 6.674\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$
Physical Concepts & Formulas
Newton’s Law of Universal Gravitation $F = GMm/r^2$ states that every mass attracts every other mass. For circular orbits, gravity provides the centripetal force: $GMm/r^2 = mv^2/r$, giving orbital speed $v = \sqrt{GM/r}$ — independent of the satellite’s mass. Kepler’s third law $T^2 \propto r^3$ follows directly. Escape velocity is found by setting kinetic energy equal to gravitational PE: $v_e = \sqrt{2GM/R}$. Inside a uniform sphere, $g$ decreases linearly with depth.
- $F = \dfrac{GMm}{r^2}$ — Newton’s Law of Gravitation
- $v_{\text{orbit}} = \sqrt{GM/r}$ — circular orbital speed
- $T = 2\pi\sqrt{r^3/GM}$ — orbital period (Kepler III)
- $v_e = \sqrt{2GM/R}$ — escape velocity from surface
- $g = GM/R^2$ — surface gravitational acceleration
Step-by-Step Solution
Step 1 — Centripetal force = Gravitational force:
$$\frac{mv^2}{r} = \frac{GMm}{r^2} \implies v = \sqrt{\frac{GM}{r}}$$
Step 2 — Period:
$$T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi\sqrt{\frac{r^3}{GM}}$$
Step 3 — Escape velocity (set $KE = PE$):
$$\frac{1}{2}mv_e^2 = \frac{GMm}{R} \implies v_e = \sqrt{\frac{2GM}{R}}$$
Worked Calculation
Substituting all values with units:
Earth: $M = 5.97\times10^{24}\,\text{kg}$, $R = 6.37\times10^6\,\text{m}$:
$$v_e = \sqrt{\frac{2\times6.674\times10^{-11}\times5.97\times10^{24}}{6.37\times10^6}} = \sqrt{\frac{7.97\times10^{14}}{6.37\times10^6}} = \sqrt{1.25\times10^8} \approx 11.2\,\text{km/s}$$
Answer
$$\boxed{v_e = \sqrt{2GM/R} \approx 11.2\,\text{km/s}}$$
Physical Interpretation
11.2 km/s ≈ 40,000 km/h is the escape speed from Earth. The Moon’s escape speed is 2.38 km/s — much lower because of its smaller mass. That is why the Moon has no atmosphere: thermal gas molecules at the Moon’s surface easily exceed this speed. Rockets do not need to reach escape velocity instantly — they can use continuous thrust, effectively integrating the Tsiolkovsky equation.
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