Weighing the Stars: Orbital Velocity Calculation Quiz

If the gravitational force (GMm/r^2) provides the centripetal force (mv^2/r), and if we solve for velocity by canceling the satellite mass 'm' and one factor of 'r', then the resulting equation is v = sqrt(GM/r).

If the centripetal force (mv^2/r) and gravitational force (GMm/r^2) both include the satellite's mass 'm', then 'm' cancels out from both sides of the equation; therefore, the velocity is independent of the satellite's mass.

If velocity 'v' is proportional to the inverse square root of the radius (1/sqrt(r)), and if the radius 'r' is multiplied by 4, then the velocity is multiplied by 1/sqrt(4), which is 1/2.

If 'G' is a fundamental constant used to calculate the force of attraction between two masses, and if standard scientific measurements are used, then its value is 6.67 x 10^-11 m^3/(kg*s^2).

If the formula for orbital velocity is v = sqrt(GM/r), then the calculation requires the central mass (M), the gravitational constant (G), and the radius (r); if the planet's mass 'm' cancels out during derivation, it is not required.

If we square both sides of v = sqrt(GM/r) to get v^2 = GM/r, and if we then multiply both sides by 'r' and divide by 'G', then the isolated variable 'M' equals v^2 * r / G.

If the orbital radius 'r' for LEO is much smaller than for a geostationary orbit, and if velocity 'v' increases as 'r' decreases (v is proportional to 1/sqrt(r)), then the LEO object must maintain a higher speed.

If centripetal acceleration is a_c = v^2/r, and if we substitute the orbital velocity squared (v^2 = GM/r), then dividing GM/r by 'r' results in a_c = GM/r^2, which is equivalent to the local gravitational field strength.

If gravity acts between the centers of mass of two bodies, then the radius 'r' must be measured from the center of the central mass, not its surface, to ensure the math follows the inverse square law.

If the gravitational constant G is measured in m^3/(kg*s^2), then all distance inputs must be in meters and mass inputs must be in kilograms so that the units cancel correctly to produce velocity in meters per second.

If velocity is distance divided by time, and if the distance for one full orbit is the circumference (2 * pi * r), then the orbital velocity is equal to (2 * pi * r) divided by the period (T).

If the formula is v = sqrt(GM/r), and if the mass 'M' is replaced with '2M', then the new velocity is sqrt(G*2M/r), which is equal to sqrt(2) multiplied by the original velocity.

If v = 2pir/T and v^2 = GM/r, then substituting for v results in (2pir/T)^2 = GM/r; if we solve for M, we get M = 4 * pi^2 * r^3 / (G * T^2), which is Kepler's Third Law.

If the variable 'r' is the distance from the center of the Earth, and if altitude is only the distance from the surface, then the Earth's radius must be added to the altitude to find the total distance from the center.

If the Sun is the central mass and v = sqrt(GM/r), then planets with the smallest 'r' (Mercury) must have the highest speed, and planets with the largest 'r' (Neptune) must have the lowest speed.

If escape velocity is the speed needed to reach zero total energy (Kinetic + Potential = 0), then v_escape = sqrt(2GM/r); if the orbital speed is v = sqrt(GM/r), then escape velocity is exactly sqrt(2) times the circular orbital speed.

If the orbital radius 'r' changes in an elliptical orbit, and if velocity is dependent on 'r' (v = sqrt(GM/r)), then the velocity must increase as the object gets closer to the focus (periapsis) and decrease as it moves away (apoapsis).

If v = sqrt(6.67x10^-11 * 2.0x10^30 / 1.5x10^11), and if we simplify the expression to sqrt(13.34x10^19 / 1.5x10^11) which is approx sqrt(8.9x10^8), then the result is roughly 29,800 m/s.

If the centripetal force used in the derivation (mv^2/r) assumes a constant radius and a direction perpendicular to the velocity, then the formula specifically describes a circular orbital path.

If the gravitational force (GMm/r^2) is what creates the acceleration, and if the acceleration (GM/r^2) is independent of the smaller mass 'm', then the speed required to balance that force depends only on the central mass 'M'.