Question

1. a geostationary satellite is moving in a circular orbit at a height of 3000 km from the earth’s surface. calculate its speed. radius of the earth, r = 6400 km
2. (a) what does the area under a stress-strain curve represent?

(b) state two examples of projectile in sports.

1. explain why the young’s modulus of aluminium is greater than that of rubber.

Explanation:

Question 3: Calculating the speed of a geostationary satellite

Step 1: Recall the formula for orbital speed

The orbital speed \( v \) of a satellite moving in a circular orbit around a planet is given by \( v = \sqrt{\frac{GM}{r}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet (Earth, in this case), and \( r \) is the radius of the orbit. However, we can also use the relation \( g = \frac{GM}{R^2} \) (where \( g \) is the acceleration due to gravity at the Earth's surface and \( R \) is the Earth's radius) to rewrite the formula for \( v \) in terms of \( g \), \( R \), and \( r \). Substituting \( GM = gR^2 \) into the orbital speed formula, we get \( v = \sqrt{\frac{gR^2}{r}} \).

First, we need to find the radius of the orbit \( r \). The satellite is at a height \( h = 3000 \, \text{km} \) from the Earth's surface, and the Earth's radius \( R = 6400 \, \text{km} \). So, \( r = R + h = 6400 \, \text{km} + 3000 \, \text{km} = 9400 \, \text{km} = 9.4 \times 10^6 \, \text{m} \). The value of \( g \) at the Earth's surface is approximately \( 9.8 \, \text{m/s}^2 \), and \( R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \).

Step 2: Substitute the values into the formula

Using \( v = \sqrt{\frac{gR^2}{r}} \), we substitute \( g = 9.8 \, \text{m/s}^2 \), \( R = 6.4 \times 10^6 \, \text{m} \), and \( r = 9.4 \times 10^6 \, \text{m} \):

\[

$$\begin{align*} v &= \sqrt{\frac{9.8 \times (6.4 \times 10^6)^2}{9.4 \times 10^6}} \\ &= \sqrt{\frac{9.8 \times 40.96 \times 10^{12}}{9.4 \times 10^6}} \\ &= \sqrt{\frac{401.408 \times 10^{12}}{9.4 \times 10^6}} \\ &= \sqrt{\frac{401.408}{9.4} \times 10^6} \\ &= \sqrt{42.703 \times 10^6} \\ &= \sqrt{4.2703 \times 10^7} \\ &\approx 6534 \, \text{m/s} \end{align*}$$

\]

(Alternatively, we can also use the formula for the speed of a geostationary satellite, but since the height is given as 3000 km (which is not the typical geostationary height, but we'll proceed with the given values), the above calculation holds. Wait, actually, a geostationary satellite is at a height of approximately 35786 km, but the problem states 3000 km, so we'll use the given height.)

Wait, maybe a simpler approach: The centripetal force for the satellite is provided by the gravitational force, so \( \frac{mv^2}{r} = \frac{GMm}{r^2} \), so \( v = \sqrt{\frac{GM}{r}} \). We know that at the Earth's surface, \( g = \frac{GM}{R^2} \), so \( GM = gR^2 \). Therefore, \( v = \sqrt{\frac{gR^2}{r}} = R\sqrt{\frac{g}{r}} \).

Substituting the values: \( R = 6400 \times 10^3 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \), \( r = (6400 + 3000) \times 10^3 \, \text{m} = 9400 \times 10^3 \, \text{m} \).

\[
v = 6400 \times 10^3 \times \sqrt{\frac{9.8}{9400 \times 10^3}}
\]

First, calculate \( \frac{9.8}{9400 \times 10^3} = \frac{9.8}{9.4 \times 10^6} \approx 1.04255 \times 10^{-6} \)

Then, \( \sqrt{1.04255 \times 10^{-6}} \approx 1.021 \times 10^{-3} \)

Then, \( v = 6400 \times 10^3 \times 1.021 \times 10^{-3} = 6400 \times 1.021 \approx 6534.4 \, \text{m/s} \approx 6.53 \times 10^3 \, \text{m/s} \) or \( 6.53 \, \text{km/s} \).

The area under a stress-strain curve for a material represents the modulus of toughness (or the total strain energy per unit volume) that the material can absorb before fracturing. It is the work done per unit volume on the material to deform it elastically and then plastically until it breaks. For a ductile material, this area is larger, indicating it can absorb more energy before failure, while for a brittle material, the area is smaller.

1. Cricket ball: When a cricket ball is bowled or hit by a bat, it follows a parabolic trajectory under the influence of gravity (after being projected), making it a projectile.
2. Basketball: When a basketball is shot towards the hoop, it is projected into the air and follows a curved path (parabola) due to gravity, thus being a projectile.

Other examples could include a football (soccer ball) when kicked for a goal attempt, a javelin in athletics, etc.

Answer:

The speed of the satellite is approximately \( \boldsymbol{6.5 \times 10^3 \, \text{m/s}} \) (or \( 6.5 \, \text{km/s} \), depending on rounding).

Question 4(a): Area under a stress-strain curve