This question asks about Kepler's Third Law of Planetary Motion, also known as the Law of Periods. This law describes the relationship between a planet's orbital period and its average distance from the Sun (or its mean orbital radius).
Correct Option: A) \(T^2\) is directly proportional to \(r^3\). This statement accurately represents Kepler's Third Law of Planetary Motion.
The question asks for the specific term describing the minimum speed required for an object to escape a planet's gravitational pull permanently without additional thrust. We need to identify the correct physics term from the given options.
B) Escape velocity is the minimum speed an object needs to break free from the gravitational attraction of a massive body without further propulsion.
To determine the escape velocity from the Earth's surface, we need to understand the concept of escape velocity and its formula. Escape velocity is the minimum speed an object needs to escape the gravitational pull of a massive body without further propulsion. It is derived by equating the kinetic energy of the object to the gravitational potential energy it needs to overcome.
Definition of Escape Velocity: Escape velocity (\(v_e\)) is the minimum velocity required for an object to escape the gravitational field of a planet or other celestial body. At this velocity, the kinetic energy of the object is equal to the magnitude of its gravitational potential energy.
Formula for Escape Velocity: The escape velocity can be calculated using the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]where:
Alternative Formula using 'g': We can also express the escape velocity in terms of the acceleration due to gravity (\(g\)) at the surface of the Earth. Since \(g = \frac{GM}{R^2}\), we have \(GM = gR^2\). Substituting this into the escape velocity formula:
\[ v_e = \sqrt{\frac{2(gR^2)}{R}} = \sqrt{2gR} \]Using the approximate values:
Calculation: Let's calculate using the simplified formula:
\[ v_e = \sqrt{2 \times 9.8 \text{ m/s}^2 \times 6.37 \times 10^6 \text{ m}} \]\[ v_e = \sqrt{124852000 \text{ m}^2/\text{s}^2} \]\[ v_e \approx 11173.7 \text{ m/s} \]Converting this to kilometers per second:
\[ v_e \approx 11.1737 \text{ km/s} \]Rounding to one decimal place, this is approximately \(11.2 \text{ km/s}\).
Correct Option: D) 11.2 km/s is the approximate escape velocity required for an object to escape the Earth's gravitational field from its surface.
The question describes a specific type of satellite that has an orbital period matching the Earth's rotational period, making it appear stationary from the Earth's surface. We need to identify the correct term for such a satellite from the given options.
B) Geostationary satellite is the correct answer. A geostationary satellite orbits the Earth at an altitude of approximately 35,786 km (22,236 miles) directly above the equator, with an orbital period of exactly one sidereal day (about 23 hours, 56 minutes, 4 seconds). This synchronized motion makes the satellite appear motionless from the ground, which is crucial for applications like telecommunications and broadcasting.
To understand what keeps an artificial satellite in orbit, we need to recall the fundamental forces acting on objects in space and the concept of centripetal force required for circular motion. A stable orbit implies a continuous balance of forces.
where \(m\) is the mass of the object, \(v\) is its orbital velocity, and \(r\) is the radius of the orbit.
where \(G\) is the gravitational constant. For a stable orbit, this gravitational force provides the exact centripetal force needed:
\[ G \frac{Mm}{r^2} = \frac{mv^2}{r} \]This balance allows the satellite to continuously "fall around" the Earth without crashing into it or flying off into space.
B) The gravitational force of attraction exerted by the Earth on the satellite. This force continuously pulls the satellite towards the Earth, acting as the centripetal force necessary to maintain its circular or elliptical orbit.
This problem involves understanding Newton's Law of Universal Gravitation, which describes the gravitational force between two objects. The key is to analyze how the force changes when the distance between the objects is altered.
C) 4 times stronger than before