Circular Reasoning
T/F: The object experiences a force which has a component directed parallel to the direction of motion.
false; if the motion is in a circle at constant speed, the net force is perpendicular to the direction of motion and there is neither a component parallel nor anti-parallel to the direction of motion.)
T/F: The gravitational force only acts between very, very massive objects
false; gravitational forces are universal (Newton's big idea); they act between any two objects which have mass.
T/F: The acceleration of gravity experienced by objects located near to (and far from) from the earth depends upon the mass of the object.
false; check out the equation - g = GMcentral/R2. The value of g does not depend upon the object's mass.
T/F: Satellites are falling projectiles.
true (mostly); satellites are projectiles which are falling towards the Earth without falling into the Earth. They are falling in the sense that (like any projectile), they are dropping below their tangential direction of motion.
Identify the type of force which causes an eraser is tied to a string swung in a horizontal circle.
a. gravity b. normal c. tension
d. applied e. friction f. spring
g. electrical h. magnetic
Tension (A string is attached to the eraser and pulls it towards the center point of the circle.)
T/F: Inertia causes objects to move in a circle.
false; it is centripetal force which causes the circular motion. Inertia (which is NOT a force) is merely the tendency of any moving object to continue in its straight-line constant speed path.
T/F: The gravitational force between an object and the earth is inversely related to the distance between the object's and the earth's center.
true; if the distance is increased, then the force is decreased.
T/F: The acceleration of gravity experienced by objects located near to (and far from) from the earth depends upon the mass of the Earth.
true; check out the equation again. The acceleration of gravity created by the earth depends upon the earth's mass.
T/F: All satellites follow circular paths.
false; some satellites travel in elliptical and even very elliptical orbits. For instance, the planets orbit the Sun in mildly elliptical paths.
Two objects attract each other with a force of gravity (Fgrav) of 44 N. If the distance separating the objects is doubled, then what is the new force of gravitational attraction?
New force = 36 N / 4 = 9.0 N
There can be a force pushing outwards on the object as long as the net force in inwards.
true; an object which moves in a circle must have a net inward force. There are many instances of individual outwards forces which are exceeded by an individual inward force
T/F: The gravitational force can ALWAYS be accurately calculated by multiplying the object mass by the acceleration of gravity (m•g).
true; this is always the case. It is not true however to say that the gravitational force is equal to mass•9.8 m/s/s. The value of g varies with location and so at distances significantly further from the earth's surface, g is reduced and the gravitational force must be computed using a different value of g.
T/F: The acceleration of gravity experienced by objects located near to (and far from) the earth is inversely related to the distance between the center of the object and the center of the earth.
true; check out the equation one more time. The separation distance is located in the denominator of the equation, indicating an inverse relationship.
T/F: The orbital velocity required of a satellite is dependent upon the mass of the satellite; a more massive satellite would require a greater orbital speed.
false; the equation for the orbital velocity of a satellite is v = SQRT(G•Mcentral/R). The Mcentral is the mass of the central body - the body being orbited by the satellite. As seen in the equation, the orbital velocity is independent of the mass of the satellite.
Determine the force of gravitational attraction between a 52.0-kg mother and a 3.0-kg child if their separation distance is 0.50 meters.
Use Newton's universal gravitation equation:
Fgrav = G • m1 • m2 /R2 = (6.67•10-11 N•m2/kg2) • (52.0 kg) • (3.0 kg) /(0.50 m)2 = 4.2•10-8 N
T/F: Because the speed is constant, the acceleration is zero.
false; acceleration occurs when there is a change in velocity. Since the direction of the velocity vector is changing, there is an acceleration - an inward acceleration.
T/F: The gravitational force acting upon an object is the same as the weight of the object.
true; weight and gravitational force are synonymous.
T/F: Increasing the mass of an object will increase the acceleration of gravity experienced by the object.
false; like statement A, this statement makes a claim that the acceleration of gravity depends upon the mass of the object. It does not. In the derivation of g from the Universal Gravitation equation, the mass cancels.
T/F: The orbital velocity of a satellite does not depend upon the mass of the planet around which it orbits.
false; the equation for the orbital velocity of a satellite is v = SQRT(G•Mcentral/R). The Mcentral is the mass of the central body - the body being orbited by the satellite. Clearly the orbital velocity depends upon the mass of the planet being orbited.
Use the following information to help determine the orbital velocity and orbital period of a satellite at a location of 15000 miles above the surface of the earth.
Mass of Earth = 5.98 x 1024 kg
Radius of Earth = 6.37 x 106 m
1609 m = 1.00 mi
Answers:
The first step is to find the separation distance from the center of the earth - this is R in most of the equations. The altitude of the satellite is 15000 mi or 2.41•107 m; this distance must be added to the radius of the earth to determine the separation distance. The radius of the earth is 6.38•106 m; the separation distance is 3.05•107 m. The orbital velocity can be computed using the orbital velocity equation:
v = SQRT(G•Mcentral/R) = SQRT[ (6.67•10-11 N•m2/kg2) • (5.98•1024 kg) / (3.05•107 m) ]
v = SQRT (1.31•107 m2/s2) = 3615 m/s
The orbital period can now be calculated using the v = 2•pi•R/T equation or the Kepler's second law equation [ T2/R3 = 4•pi2/(GMearth)]. Starting with the first equation, algebra yields:
T = (2•pi•R)/v = [2 • pi • (3.05•107 m) / (3615 m/s) ] = 5.30 • 104 s = 14.7 hr
T/F: The acceleration and the net force vector are directed perpendicular to each other
false; the acceleration and net force are always directed in the same direction. In this case, F and a are directed inward; this happens to be perpendicular to the tangential velocity vector.
T/F: The gravitational force between two objects is independent of the mass of the smaller of the two objects.
false; gravitational force is dependent upon the product of the two masses. Both masses are important in the computation
T/F: Doubling the distance between an object and the earth's center will decrease the acceleration of gravity by a factor of four.
true; g is inversely proportional to the square of the distance; a doubling of the distance means that you must divide the force of gravity value by 4 (22) to obtain the new force of gravity value.
T/F: A high-altitude satellite will require a greater orbital speed than a low-altitude satellite.
false the equation for the orbital velocity of a satellite is v = SQRT(G•Mcentral/R). The R in the denominator inside the radical is the radius of the orbit. The higher altitude satellites have a greaer radius of orbit. As the R in the denominator increases, the required orbital velocity will decrease.
Distinguish between true- and apparent-weightlessness.
True weightlessness occurs when there is an absence of gravitational force on an object. This is not a likely occurrence since every mass in the universe is attracted to every other mass in the universe with some force of gravity. All objects would experience some degree of gravitational attraction (though it may be considerably small). Apparent weightlessness is the sensation of not experiencing any external contact forces. This occurs when an object is in free-fall.