An earth satellite describe a circular orbit of radius R about center of the earth which is assumed spherical. Calculate the value of R if the period of the orbit is equal to one day

Gravitational force:

$$F_g = \frac{G M m}{R^2}$$

Centripetal force:

$$F_c = \frac{m v^2}{R}$$

Set $F_g = F_c$ (both in N or kg $\cdot$ m/s$^2$):

$$\frac{G M m}{R^2} = \frac{m v^2}{R}$$

Cancel $m$:

$$\frac{G M}{R^2} = \frac{v^2}{R}$$

Multiply both sides by $R$:

$$\frac{G M}{R} = v^2 \quad (\mathrm{m^2/s^2})$$

$$v = \frac{2 \pi R}{T} \quad (\mathrm{m/s})$$

Substitute into $v^2$:

$$\frac{G M}{R} = \left( \frac{2 \pi R}{T} \right)^2 = \frac{4 \pi^2 R^2}{T^2}$$

Multiply both sides by $R T^2$:

$$G M T^2 = 4 \pi^2 R^3$$ $$R^3 = \frac{G M T^2}{4 \pi^2} \quad (\mathrm{m^3})$$ $$R = \left( \frac{G M T^2}{4 \pi^2} \right)^{1/3}$$

Period: $T = 1$ day $= 24 \times 60 \times 60 = 86400$ s

Earth's gravitational parameter: $G M = 3.986 \times 10^{14}$ m$^3$ s$^{-2}$

Compute $T^2$:

$$T^2 = (86400)^2 = 7.46496 \times 10^9 \ \mathrm{s^2}$$

Compute $G M T^2$:

$$G M T^2 = 3.986 \times 10^{14} \times 7.46496 \times 10^9 = 2.9755 \times 10^{24} \ \mathrm{m^3}$$

Compute $4 \pi^2$:

$$\pi^2 \approx 9.8696, \quad 4 \pi^2 \approx 39.4784$$

Compute fraction:

$$\frac{G M T^2}{4 \pi^2} = \frac{2.9755 \times 10^{24}}{39.4784} = 7.538 \times 10^{22} \ \mathrm{m^3}$$

Compute cube root:

$$R = \sqrt[3]{7.538 \times 10^{22}} = 4.22 \times 10^7 \ \mathrm{m}$$

Verification: $(4.22 \times 10^7)^3 = 4.22^3 \times 10^{21} \approx 75.1 \times 10^{21} = 7.51 \times 10^{22} \ \mathrm{m^3}$ (matches closely).

4.22 \times 10^{7} \ m