Forces (2) Flashcards

Question

Radial Vs Uniform field lines

Answer 1

**Radial Fields**: - **Non-uniform**, with **field strength (g)** varying with distance from the center. - Represented by **radially inward lines**. === **Uniform Fields**: - Have **equally spaced parallel lines**, with **field strength (g)** remaining constant. - On the **Earth’s surface**, gravitational field lines are **parallel**, approximating a **uniform field**.

Answer 2

**Newton’s Law of Gravitation**: - The **gravitational force (F)** is: - **Directly proportional** to the product of two masses (**m₁** and **m₂**). - **Inversely proportional** to the square of their separation (**r**). - The **negative sign** indicates the force is **attractive**. === **Distance (r)**: - Measured between the **centers** of the two masses. - For objects above a planet’s surface, add the **planet’s radius** to the height above the surface.

Answer 3

**Gravitational Field Strength (g) in a Radial Field**: - **Equation**: **g = -GM/r²**, where: - G = gravitational constant. - M = mass of the object. - r = distance from the center of the mass. - The **negative sign** indicates the field is **attractive**, with field lines pointing toward the mass. === **Key Points**: - **g** decreases with increasing **r**, following an **inverse square law**. - On **Earth’s surface**, g ≈ 9.81 N kg⁻¹, but outside Earth, **g** is **not constant**.

Answer 4

- **Definition**: Work done per unit mass to move a test mass from **infinity** to a point in the field. - **Key Properties**: - **Always negative** (attractive force). - **Zero at infinity** (reference point). - **Less negative** with increasing \(r\). === **2. Gravitational Potential Energy (\(E_p\))** - **Definition**: Energy of a mass \(m\) due to its position in a gravitational field. - **Key Points**: - **Increases** (becomes less negative) as \(r\) increases. - **Maximum \(E_p = 0\)** at infinity.

Answer 5

- **Gravitational potential difference (ΔV)** is the **difference in gravitational potential between two** points at different distances from a mass. - **ΔV** is given by **ΔV = Vf – Vi**, where **Vf** is the final potential and **Vi** is the initial potential.

Answer 6

**Escape Velocity**: - The **minimum speed** required for an object to escape a **gravitational field** without further energy input. - Depends on the **mass (M)** and **radius (r)** of the object creating the gravitational field. - **Same for all objects** in the same gravitational field. === **Equation**: - **v = √(2GM/r)**, where: - v = escape velocity. - G = gravitational constant. - M = mass of the object. - r = distance from the object’s center. === **Key Point**: - An object reaches **escape velocity** when all its **kinetic energy** is transferred to **gravitational potential energy**.

Answer 7

**Satellite Motion**: - The **gravitational force** on a satellite acts as the **centripetal force**, always perpendicular to the direction of travel. === **Orbital Speed Equation**: - Equating **gravitational force** to **centripetal force** gives: **v = √(GM/r)**, where: - v = orbital speed. - G = gravitational constant. - M = mass of the central object. - r = distance from the center of the object to the satellite. === **Key Point**: - The **mass of the satellite (m)** cancels out, showing that **orbital speed** is independent of the satellite’s mass.

Answer 8

- The **orbital speed** 𝑣 of a planet or satellite in a circular orbit is **given by the circular motion formula v = 2π/T** - **v² = GM/R** - **Equal the two and solve for T²**

Answer 9

**Force-Distance Graph for Gravitational Fields**: - The graph is a **curve** because **force (F)** is inversely proportional to **r²**. - The **area under the graph** represents **work done** or **energy transferred**. === **Gravitational Potential Energy**: - When a mass **m** moves further from mass **M**, its **gravitational potential energy** increases, requiring **work** to be done. - The **area between two points** on the graph gives the **change in gravitational potential energy** of mass **m**.

Answer 10

**Geostationary Orbit**: - A satellite remains **directly above the equator**, in the plane of the equator, and always at the same point above Earth’s surface. - The satellite **moves west to east** with an **orbital period of 24 hours**, matching Earth’s rotation. === **Applications**: - Used for **telecommunication transmissions** and **television broadcasts**, where the satellite amplifies and sends signals back to Earth. - **Receiver dishes** on Earth can be **fixed**, as they always point to the same spot in the sky due to the satellite’s fixed position.

Answer 11

**Kepler’s First Law**: - The orbit of a planet is an **ellipse** with the Sun at one of the two foci. - Orbits can be **circular** or **highly elliptical**. === **Kepler’s Second Law**: - A line joining a planet and the Sun sweeps out **equal areas in equal time intervals**. - Planets move **faster** when closer to the Sun and **slower** when farther away. === **Kepler’s Third Law**: - The **square of the orbital period (T)** is directly proportional to the **cube of the orbital radius (r)**: **T² ∝ r³** - A **constant (k)** applies to all planets in the same system.

Answer 12

**Circular Motion in a Vertical Loop**: - **Bottom of the Loop**: - Centripetal Force (F) = **R - mg** (difference between upward force **R** and weight **mg**). - **Top of the Loop**: - Centripetal Force (F) = **R + mg** (sum of upward force **R** and weight **mg**). - **Side of the Loop**: - Centripetal Force (F) = **R** (equal to the upward force). === **Circular Motion on a Banked Track**: - **Forces**: - **Weight (mg)**: Acts vertically downward. - **Normal Force (N)**: Acts perpendicular to the surface. - **Components of Normal Force (N)**: - **Vertical**: Balances weight. - **Horizontal**: Provides centripetal force. - **Angle of Banking (θ)**: Determines the ratio of the normal force’s components, reducing reliance on friction.

Answer 13

**Height Difference in Pendulum Motion**: - The **height difference** is **not** the amplitude. - Use **mgh = KE** to relate height to kinetic energy. - **Maximum velocity (vₘₐₓ)** can be found using the energy equation.

Answer 14

Less on equator because some of the weight is used as centripetal force

Forces (2) Flashcards

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