Calculate the orbital speed and period of an object in a circular orbit around Earth at an altitude of 230 km.

The orbital radius is the sum of the Earth's radius and the altitude.

$$r = R_E + h$$ $$r = 6.371 \times 10^6 \text{ m} + 2.30 \times 10^5 \text{ m}$$ $$r = 6.371 \times 10^6 \text{ m} + 0.230 \times 10^6 \text{ m}$$ $$r = (6.371 + 0.230) \times 10^6 \text{ m}$$ $$r = 6.601 \times 10^6 \text{ m}$$

For an object in a circular orbit, the gravitational force provides the centripetal force.

$$\frac{GM_E m}{r^2} = \frac{mv^2}{r}$$

Solving for $v$:

$$v = \sqrt{\frac{GM_E}{r}}$$

Substitute the values:

$$v = \sqrt{\frac{(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (5.972 \times 10^{24} \text{ kg})}{6.601 \times 10^6 \text{ m}}}$$ $$v = \sqrt{\frac{3.9833 \times 10^{14} \text{ m}^3/\text{s}^2}{6.601 \times 10^6 \text{ m}}}$$ $$v = \sqrt{6.0344 \times 10^7 \text{ m}^2/\text{s}^2}$$ $$v \approx 7768.1 \text{ m/s}$$

Rounding to three significant figures:

$$v \approx 7770 \text{ m/s}$$

The speed in orbit is $\boxed{7770 \text{ m/s}}$.

2. Find the time for one complete orbit.

The time for one complete orbit (period) is the circumference of the orbit divided by the orbital speed.

$$T = \frac{2\pi r}{v}$$

Substitute the values:

$$T = \frac{2\pi (6.601 \times 10^6 \text{ m})}{7768.1 \text{ m/s}}$$ $$T = \frac{4.1474 \times 10^7 \text{ m}}{7768.1 \text{ m/s}}$$ $$T \approx 5339.2 \text{ s}$$

Rounding to three significant figures:

$$T \approx 5340 \text{ s}$$

To convert to minutes:

$$T = 5339.2 \text{ s} \times \frac{1 \text{ min}}{60 \text{ s}}$$ $$T \approx 88.987 \text{ min}$$

Rounding to three significant figures:

$$T \approx 89.0 \text{ min}$$

The time for one complete orbit is $\boxed{5340 \text{ s}}$ or $\boxed{89.0 \text{ min}}$.