Physics Chapter on Angle and Velocity

<p>m s-2</p> Signup and view all the answers

<p>Force required to maintain uniform circular motion towards the center</p> Signup and view all the answers

<p>Force is directly proportional to the product of masses and inversely proportional to the square of distance</p> Signup and view all the answers

<p>Vector quantity</p> Signup and view all the answers

<p>Centripetal acceleration</p> Signup and view all the answers

<p>s = θr</p> Signup and view all the answers

<p>T^2 = (4π^2 R^3) / GM</p> Signup and view all the answers

<p>They are falling at the same rate as their spacecraft.</p> Signup and view all the answers

<p>3.8 × 10^8 m</p> Signup and view all the answers

<p>Both B and C</p> Signup and view all the answers

<p>36,000 km</p> Signup and view all the answers

<p>It maintains a circular motion compensating for gravitational pull.</p> Signup and view all the answers

<p>It is directly proportional to the product of the masses.</p> Signup and view all the answers

<p>T^2 is directly proportional to R^3</p> Signup and view all the answers

<p>0.57 m/s²</p> Signup and view all the answers

<p>Angular velocity ($,\omega$) is defined as the angle rotated ($\theta$) divided by the periodic time ($T$), expressed as $\omega = \frac{\theta}{T}$.</p> Signup and view all the answers

<p>Centripetal acceleration is determined by the square of the object's speed ($v^2$) divided by the radius ($r$) of the circular path, given by the formula $a_c = \frac{v^2}{r}$.</p> Signup and view all the answers

<p>Centripetal force is necessary to keep an object moving in a circular path; it acts towards the center of the circle and is calculated as $F_c = \frac{mv^2}{r}$, where $m$ is mass and $v$ is linear velocity.</p> Signup and view all the answers

<p>The orbital speed ($v$) of a satellite can be expressed as $v = \sqrt{\frac{GM}{r}}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth.</p> Signup and view all the answers

<p>Newton's Universal Law of Gravitation states that the force of attraction between two masses is directly proportional to their mass and inversely proportional to the square of the distance between them, which is fundamental for understanding how objects orbit.</p> Signup and view all the answers

<p>The orbital period can be derived using the formula $T^2 = \frac{4\pi^2 R^3}{GM}$, where $T$ is the period, $R$ is the radius of the orbit, $G$ is the gravitational constant, and $M$ is the mass of the central body.</p> Signup and view all the answers

<p>Geostationary satellites orbit at about 36,000 km above the Earth with an orbital period of 24 hours, matching the Earth's rotation speed, allowing them to appear stationary above a specific point.</p> Signup and view all the answers

<p>The gravitational force acting on the satellite is equal to the centripetal force needed to keep it in circular motion, given by the equation $F_{gravitational} = F_{centripetal}$.</p> Signup and view all the answers

<p>The average radius is critical as it helps determine the gravitational force and the orbital period of the object, as shown in Kepler’s laws and gravitational equations.</p> Signup and view all the answers

<p>Astronauts feel weightless because both they and their spacecraft are in free fall towards Earth at the same rate, creating a state of constant acceleration without a normal force acting on them.</p> Signup and view all the answers

<p>By equating the gravitational force <code>GMm/R^2</code> and the centripetal force <code>mv^2/R</code>, we can isolate <code>M</code>, leading to the formula <code>M=(v^2 R)/G</code>.</p> Signup and view all the answers

<p>Astronauts appear weightless because both they and their spacecraft are in free fall, accelerating towards Earth at the same rate.</p> Signup and view all the answers

<p>Kepler’s Third Law states that the square of the orbital period <code>T^2</code> is proportional to the cube of the average radius <code>R^3</code> of the orbit.</p> Signup and view all the answers

<p>Geostationary satellites are positioned at approximately 36,000 km altitude to match the Earth's rotation period, allowing them to remain stationary relative to the surface.</p> Signup and view all the answers

<p>The gravitational force provides the necessary centripetal force, while the satellite's horizontal velocity counters this pull, enabling it to travel in a stable circular path.</p> Signup and view all the answers

<p>Angular velocity is related to linear velocity by the equation $v = heta r$.</p> Signup and view all the answers

<p>Centripetal force acts toward the center of the circle, maintaining an object's circular motion by continuously changing its direction.</p> Signup and view all the answers

<p>Periodic time is the inverse of angular velocity, given by the equation $T = \frac{2\pi}{\omega}$.</p> Signup and view all the answers

<p>Centripetal acceleration is a vector quantity because it has both direction and magnitude, always pointing towards the center of circular motion.</p> Signup and view all the answers

<p>The gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers.</p> Signup and view all the answers