Problem 1.46 — Satellite orbital speed

Problem Statement

Solve the gravitation problem: Find the orbital speed and period of a satellite at height $h = 200\,\text{km}$ above the Earth’s surface. ($R_E = 6370\,\text{km}$, $g = 9.8\,\text{m/s}^2$) Orbital radius: $r = R_E+h = 6570\,\text{km} = 6.57\times10^6\,\text{m}$ Gravitational acceleration at altitude: $$g(r) = g\frac{R_E^2}{r^2} =

Given Information

• $h = 200\,\text{km}$
• $R_E = 6370\,\text{km}$
• $g = 9.8\,\text{m/s}$
• $r = R_E+h = 6570\,\text{km}$

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

• See the step-by-step solution for the specific equations applied.
• All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$g(r) = g\frac{R_E^2}{r^2} =

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:

$$

$$

Step 2 — Period:

$$

$$

Step 3 — Escape velocity (set $KE = PE$):

$$

Answer

$$\boxed{v_e = \sqrt{2GM/R} \approx 11.2\,\text{km/s}}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.