Space Exit Ticket
| # | Answer |
|---|---|
| 1 | Gravity. |
| 2 | Planets move in elliptical orbits with the Sun at one focus. |
| 3 | Escape velocity is the minimum speed an object must have to leave a celestial body’s gravitational field without returning, assuming no additional propulsion and ignoring air resistance. |
| 4 | It is in continuous free fall around Earth, so both the spacecraft and the object accelerate together and there is no support force. |
| 5 | It decreases. Higher orbits have lower required orbital speed because gravity is weaker at larger distances from Earth. |
| 6 | Gravitational force is inversely proportional to the square of the distance between the objects, F ∝ 1/r^2. |
| 7 | Surface gravity depends on mass and radius. The Moon’s much smaller mass means it produces a weaker gravitational field at its surface. |
| 8 | Gravity provides the centripetal force. |
| 9 | The orbital period increases. |
| 10 | Gravity is still significant in orbit, but the astronauts are in free fall, so they experience very small apparent weight, not zero gravity. |
| 11 | It becomes one-fourth as large, because surface gravity is proportional to M/r^2 and the radius is doubled. |
| 12 | The planet’s gravitational field strength at that location determines the acceleration. |
| 13 | Rockets can produce thrust in space by carrying both fuel and oxidizer, and they can reach the very high speeds needed to overcome gravity and reach orbit; airplanes require air for lift and combustion. |
| 14 | Set gravitational force equal to centripetal force: GMm/r^2 = mv^2/r. |
| 15 | Its orbital speed does not change, because orbital speed depends on the central body and orbital radius, not on the satellite’s mass. |
| 16 | Its gravitational potential energy increases (becomes less negative) because it is farther from Earth and less tightly bound gravitationally. |
| 17 | Gravity exists throughout space; objects in space still experience gravitational attraction from planets, stars, and other masses. Spacecraft feel weightless mainly because they are in free fall. |
| 18 | F = Gm1m2/r^2 = (6.67 × 10^-11)(5.0)(5.0)/(2.0^2) ≈ 4.2 × 10^-10 N. |
| 19 | g = GM/r^2 = (6.67 × 10^-11)(6.0 × 10^24)/(3.0 × 10^6)^2 ≈ 44.4 m/s^2. |
| 20 | It will escape the planet and still have excess kinetic energy far away. Since escape velocity is the minimum speed needed to reach infinity with zero final speed, a launch speed of 2v means the probe leaves with positive total energy and will move away indefinitely. |