Question

global positioning systems (gps) satellite navigation system, which tells drivers their exact position and the best route to their destination. there is a network of global positioning satellites that orbit earth at an altitude of (2.01 \times 10^7\text{ m}).

1. the orbital period of these satellites is

a. (7.88\text{ h})
b. (11.9\text{ h})
c. (24.0\text{ h})
d. (713\text{ h})

use the following information to answer the next question.
a (300\text{ g}) ball is swinging in a vertical circle on the end of a (1.20\text{ m}) string. the speed of the ball at the bottom of the swing is (8.50\text{ m/s}).

1. the tension in the string is

a. (2.94\text{ n})
b. (15.1\text{ n})
c. (18.1\text{ n})
d. (21.0\text{ n})

1. a force of (40.0\text{ n}) is applied to a rope that makes an angle of (60.0^\circ) to the object attached to the rope moves horizontally along a frictionless surface for (3.00\text{ m}). how much work is done?

Explanation:

Calculate orbital period of GPS satellites

The altitude of the GPS satellites is given as \(h = 2.01 \times 10^7\text{ m}\).
The radius of the Earth is \(R_E \approx 6.37 \times 10^6\text{ m}\).
The orbital radius is:

$$r = R_E + h = 6.37 \times 10^6 + 2.01 \times 10^7 = 2.647 \times 10^7\text{ m}$$

Using Kepler's Third Law or orbital velocity:

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

where \(G = 6.674 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2\) and \(M = 5.972 \times 10^{24}\text{ kg}\).

$$v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{2.647 \times 10^7}} \approx 3880\text{ m/s}$$

The orbital period \(T\) is:

$$T = \frac{2\pi r}{v} = \frac{2\pi \times 2.647 \times 10^7}{3880} \approx 42860\text{ s}$$

Convert to hours:

$$T = \frac{42860}{3600} \approx 11.9\text{ h}$$

Determine tension at the bottom of a vertical circle

The mass of the ball is \(m = 300\text{ g} = 0.300\text{ kg}\).
The radius of the circle is \(r = 1.20\text{ m}\).
The speed at the bottom is \(v = 8.50\text{ m/s}\).
At the bottom of the swing, the forces acting on the ball are tension \(T\) upward and gravity \(mg\) downward.
The net centripetal force is:

$$F_c = T - mg = \frac{mv^2}{r}$$

Solving for tension \(T\):

$$T = mg + \frac{mv^2}{r}$$

$$T = (0.300 \times 9.81) + \frac{0.300 \times 8.50^2}{1.20}$$

$$T = 2.943 + 18.0625 = 21.0055\text{ N} \approx 21.0\text{ N}$$

Calculate work done by the applied force

The applied force is \(F = 40.0\text{ N}\).
The angle with the horizontal is \(\theta = 60.0^\circ\).
The displacement is cut off, but we can analyze the formula for work:

$$W = F d \cos\theta$$

Since the displacement \(d\) is not fully visible in the image, we solve the first two fully visible multiple-choice questions.

Answer:

Question 1

• a. 7.88 h
• b. 11.9 h (Correct answer)
• c. 24.0 h
• d. 713 h

Question 2

• a. 2.94 N
• b. 15.1 N
• c. 18.1 N
• d. 21.0 N (Correct answer)