Problem Statement
Find the radius $r_0$ of a geostationary orbit (period 24 hours).
Given Information
- See problem statement for all given quantities.
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 — Identify given quantities and set up the problem: For a geostationary orbit: $T = 24\,\text{h} = 86400\,\text{s}$, gravity provides centripetal force:
Step 2 — Apply the relevant physical law or equation: $$\frac{g R_E^2}{r_0^2} = \frac{(2\pi r_0)^2}{T^2 r_0} = \frac{4\pi^2 r_0}{T^2}$$
$$r_0^3 = \frac{gR_E^2 T^2}{4\pi^2}$$
$$r_0 = \left(\frac{gR_E^2 T^2}{4\pi^2}\right)^{1/3}$$
Step 3 — Solve algebraically for the unknown: Numerically:
Step 4 — Substitute numerical values with units: $$r_0 = \left(\frac{9.8\times(6.37\times10^6)^2\times(86400)^2}{4\pi^2}\right)^{1/3}$$
$$= \left(\frac{9.8\times4.06\times10^{13}\times7.46\times10^9}{39.5}\right)^{1/3}$$
$$= \left(7.54\times10^{22}\right)^{1/3} \approx 4.22\times10^7\,\text{m} \approx \boxed{42200\,\text{km}}$$
Worked Calculation
$$\frac{g R_E^2}{r_0^2} = \frac{(2\pi r_0)^2}{T^2 r_0} = \frac{4\pi^2 r_0}{T^2}$$
$$r_0^3 = \frac{gR_E^2 T^2}{4\pi^2}$$
$$r_0 = \left(\frac{gR_E^2 T^2}{4\pi^2}\right)^{1/3}$$
For a geostationary orbit: $T = 24\,\text{h} = 86400\,\text{s}$, gravity provides centripetal force:
$$\frac{g R_E^2}{r_0^2} = \frac{(2\pi r_0)^2}{T^2 r_0} = \frac{4\pi^2 r_0}{T^2}$$
$$r_0^3 = \frac{gR_E^2 T^2}{4\pi^2}$$
$$r_0 = \left(\frac{gR_E^2 T^2}{4\pi^2}\right)^{1/3}$$
Numerically:
$$r_0 = \left(\frac{9.8\times(6.37\times10^6)^2\times(86400)^2}{4\pi^2}\right)^{1/3}$$
$$= \left(\frac{9.8\times4.06\times10^{13}\times7.46\times10^9}{39.5}\right)^{1/3}$$
$$= \left(7.54\times10^{22}\right)^{1/3} \approx 4.22\times10^7\,\text{m} \approx \boxed{42200\,\text{km}}$$
Answer
$$= \left(7.54\times10^{22}\right)^{1/3} \approx 4.22\times10^7\,\text{m} \approx \boxed{42200\,\text{km}}$$
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.
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