Please answer the question . The answers should be in English please

The gravitational force provides the centripetal force for the satellite. The gravitational field strength $g'$ at a distance $r_s$ from the center of the Earth is related to the surface gravitational field strength $g$ by the formula:

$$g' = g \left(\frac{R}{r_s}\right)^2$$

The centripetal force $F_c$ acting on the satellite is then $F_c = m g'$.
Substituting $r_s = 5.7R$:

$$F_c = m \cdot g \left(\frac{R}{5.7R}\right)^2 = m \cdot g \left(\frac{1}{5.7}\right)^2 = \frac{m g}{(5.7)^2}$$

$$F_c = \frac{60 \text{ kg} \times 9.8 \text{ N/kg}}{(5.7)^2}$$ $$F_c = \frac{588}{32.49}$$ $$F_c \approx 18.09786 \text{ N}$$

Rounding to two significant figures:

$$F_c \approx \boxed{\text{18 N}}$$

$$r_s = 5.7R = 5.7 \times (6.4 \times 10^6 \text{ m})$$ $$r_s = 3.648 \times 10^7 \text{ m}$$

The centripetal force is also given by $F_c = \frac{m v^2}{r_s}$. Equating this to the gravitational force $m g'$:

$$\frac{m v^2}{r_s} = m g'$$ $$v^2 = g' r_s$$

Substitute $g' = g \left(\frac{R}{r_s}\right)^2$:

$$v^2 = g \left(\frac{R}{r_s}\right)^2 r_s = \frac{g R^2}{r_s}$$ $$v = \sqrt{\frac{g R^2}{r_s}}$$

Substitute the values:

$$v = \sqrt{\frac{9.8 \text{ m/s}^2 \times (6.4 \times 10^6 \text{ m})^2}{3.648 \times 10^7 \text{ m}}}$$ $$v = \sqrt{\frac{9.8 \times 40.96 \times 10^{12}}{3.648 \times 10^7}}$$ $$v = \sqrt{1.09999 \times 10^7 \text{ m}^2/\text{s}^2}$$ $$v \approx 3316.6 \text{ m/s}$$

The period $T$ is the time taken for one complete orbit, given by $T = \frac{2 \pi r_s}{v}$.

$$T = \frac{2 \pi \times (3.648 \times 10^7 \text{ m})}{3316.6 \text{ m/s}}$$ $$T \approx \frac{229219440}{3316.6}$$ $$T \approx 69110.8 \text{ s}$$

Rounding to two significant figures:

$$T \approx \boxed{6.9 \times 10^4 \text{ s}}$$