Important Practice Mid-Term Exam Questions

Exam Practice Questions on Celestial Mechanics and Laws of Planetary Motion and Gravity and Orbits Question 1: Kepler's Laws of Planetary Motion Part A: State Kepler's three laws of planetary motion. Answer: 1. First Law (Law of Ellipses): - Circles or ellipses are orbits in which the Sun lies at one focus of each. 2. Second Law (Law of Equal Areas): - If we draw a line segment connecting a planet and the Sun, then a shape called an arc, the area of that shape changes proportionally to the area that the planet takes up over equal fractional parts of the orbit. 3. Third Law (Law of Harmonies) - Orbital period of a planet is defined to be the time required by that planet to complete an orbit around the sun, and is denoted as T, while the semi-major axis is defined as the average distance of a planet from the sun and is denoted as a. Part B: Explain how Kepler’s second law (the law of equal areas) applies to the speed of a planet in its orbit around the Sun. Answer: This makes the speeds of planets to be higher when they are near to the Sun and less when far from the Sun called Speed Variation. Secondly, Area Swept because The area of space which the line joining the planet and the sun covers in equal intervals of time is equal; this is evidence of varying speed. Question 2: Newton's Law of Universal Gravitation Part A: Write down the equation for Newton's law of universal gravitation and define each term. Answer: ● ● ● ● F: Gravitational force between two objects G: Gravitational constant (6.674×10^-11 N(m/kg)^2) m1: Mass of the first object m2​: Mass of the second object ● r: Distance between the centers of the two objects Part B: How does the gravitational force between two objects change if the distance between them is doubled? Answer: Evidently, the force of gravity is inversely in relation to the square of the distance as postulated by the Inverse Square Law. Hence, it may be observed that with the increase in distance 2r, the force F is only a quarter of the force that was earlier experienced. Question 3: Orbital Mechanics Part A: Describe the difference between geostationary and polar orbits. A geostationary orbit is located about 35786 kilometers over the equatorial plane and take a time of 24 hours to complete one revolution around the earth. This makes the satellite become fixed in relation to one point on the equator hence is widely used in communication and weather satellites. An example of such an orbit is a polar orbit which goes over the poles of the Earth, and the satellite passes over the entire surface of the Earth one time. Most of these orbits are between 200 to 1000 Km, the orbiting period is from 90 to 120 min. Polar orbits are particularly useful in earth observation, reconnaissance as well as environmental surveillance. Part B: Calculate the orbital period of a satellite orbiting Earth at an altitude of 300 km. Assume the Earth's radius is 6371 km and the gravitational constant is 6.674×10-11 N(m/kg)2. The start point in the procedure of calculating the orbital period is to establish the orbital radius: A satellite orbiting the Earth at an altitude of 300 kilometers from the Earth surface. Thus, the Earth’s radius of 6,371 kilometers added to an altitude of 300 kilometers will yield an orbital radius of 6,671, or 6. 671 × 10^6 meters. The force of gravity which is the centripetal force needed for circular motion is described by Newton’s product of gravitation masses and inverse of the square of distance. Substituting the gravitational force with the centripetal force, we have fill in; orbital velocity v equals RGM with G as gravitational constant equal to 6. 674×10^-11 N(m/kg)^2, M as the mass of the Earth equal to 5. 972×10^24 kg. Having substituted the values we obtain the orbital velocity to be approximately 7. 7. 3 x 10^3 m/s. T≈5424 seconds, where 90 is round about. 4 minutes.