3. 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
4. (a) what does the area under a stress - strain curve represent? (b) state two examples of projectile in sports.
5. explain why the young’s modulus of aluminium is greater than that of rubber.

Question

Accepted Answer

### Problem 3: Geostationary Satellite Speed Calculation

# Explanation:
## Step 1: Identify the formula for orbital speed
The orbital speed $ v $ of a satellite moving in a circular orbit around a planet is given by the formula derived from equating gravitational force to centripetal force:
$ v = \sqrt{\frac{GM}{r}} $, where $ G $ is the gravitational constant, $ M $ is the mass of the Earth, and $ r $ is the radius of the orbit.

Alternatively, using 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), we can rewrite the orbital speed formula as:
$ v = \sqrt{\frac{gR^2}{r}} $.

We know $ g \approx 9.8 \, \text{m/s}^2 $, $ R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} $, and the orbital radius $ r = R + h $, where $ h = 3000 \, \text{km} = 3 \times 10^6 \, \text{m} $.

## Step 2: Calculate the orbital radius $ r $
$ r = R + h = 6.4 \times 10^6 \, \text{m} + 3 \times 10^6 \, \text{m} = 9.4 \times 10^6 \, \text{m} $.

## Step 3: Substitute values into the orbital speed formula
Using $ v = \sqrt{\frac{gR^2}{r}} $:
First, calculate $ gR^2 $:
$ gR^2 = 9.8 \, \text{m/s}^2 \times (6.4 \times 10^6 \, \text{m})^2 = 9.8 \times 40.96 \times 10^{12} \, \text{m}^3/\text{s}^2 = 401.408 \times 10^{12} \, \text{m}^3/\text{s}^2 $.

Then divide by $ r $:
$ \frac{gR^2}{r} = \frac{401.408 \times 10^{12} \, \text{m}^3/\text{s}^2}{9.4 \times 10^6 \, \text{m}} \approx 42.703 \times 10^6 \, \text{m}^2/\text{s}^2 $.

Finally, take the square root:
$ v = \sqrt{42.703 \times 10^6 \, \text{m}^2/\text{s}^2} \approx \sqrt{4.2703 \times 10^7} \, \text{m/s} \approx 6534 \, \text{m/s} $ (or $ \approx 6.53 \, \text{km/s} $).

# Answer:
The speed of the geostationary satellite is approximately $ \boldsymbol{6.5 \, \text{km/s}} $ (or $ 6500 \, \text{m/s} $ to $ 6550 \, \text{m/s} $ depending on precision).

### Problem 4(a): Area Under Stress-Strain Curve

# Brief Explanations:
The area under a stress-strain curve (for a material) represents the **work done per unit volume** to deform the material up to the point corresponding to the end of the area (e.g., up to the elastic limit, yield point, or fracture point). In the elastic region, it represents the elastic strain energy stored per unit volume (resilience). For the entire curve (up to failure), it represents the total energy absorbed per unit volume before fracture (toughness).

# Answer:
The area under a stress-strain curve represents the **work done per unit volume** to deform the material (or the energy absorbed per unit volume, such as resilience in the elastic region or toughness up to fracture).

### Problem 4(b): Examples of Projectile in Sports

# Brief Explanations:
A projectile is an object launched into the air with an initial velocity, moving under gravity (and air resistance, often neglected). In sports:
1. **Cricket Ball**: When a batsman hits the ball, it follows a parabolic trajectory (projectile motion) until it lands or is caught.
2. **Basketball Shot**: When a player shoots the ball, it is projected into the air towards the hoop, moving in a projectile path.

# Answer:
Two examples of projectiles in sports:
1. A cricket ball after being hit by a bat.
2. A basketball during a shot towards the hoop.

### Problem 5: Young’s Modulus of Aluminium vs. Rubber

# Brief Explanations:
Young’s modulus ($ Y $) is a measure of a material’s stiffness: $ Y = \frac{\text{Stress}}{\text{Strain}} $ (for elastic deformation). A higher $ Y $ means the material is stiffer (requires more stress to produce the same strain).

- **Aluminium**: It is a metal with strong interatomic bonds. When stress is applied, the atoms resist deformation strongly, so a small strain is produced for a given stress.
- **Rubber**: It is an elastomer with weak intermolecular forces and flexible molecular chains. Even a small stress causes large deformation (high strain).

Since $ Y = \frac{\text{Stress}}{\text{Strain}} $, for the same stress, aluminium has much smaller strain than rubber. Thus, $ Y_{\text{Aluminium}} > Y_{\text{Rubber}} $.

# Answer:
Young’s modulus of aluminium is greater than that of rubber because aluminium (a metal) has strong interatomic bonds, so it resists deformation (has small strain) under stress. Rubber (an elastomer) has weak intermolecular forces and flexible molecular chains, so it deforms easily (has large strain) under the same stress. Since $ Y = \frac{\text{Stress}}{\text{Strain}} $, a smaller strain for the same stress gives a higher $ Y $ for aluminium.

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QUESTION IMAGE

Question

Explanation:

Problem 3: Geostationary Satellite Speed Calculation

Step 1: Identify the formula for orbital speed

Step 2: Calculate the orbital radius \( r \)

Step 3: Substitute values into the orbital speed formula

Answer:

Problem 4(a): Area Under Stress-Strain Curve