Physics: Gravitational Fields and Orbital Mechanics
Calculating gravitational force, field strength, orbital speed, and period
Physics: Gravitational Fields and Orbital Mechanics
Calculating gravitational force, field strength, orbital speed, and period
Physics - Grade 9-12
- 1
Two students model gravity using two spheres. Sphere A has a mass of 4.0 kg, sphere B has a mass of 6.0 kg, and their centers are 0.50 m apart. Calculate the gravitational force between them.
Use Newton's law of universal gravitation, F = Gm1m2/r^2.
The gravitational force is F = Gm1m2/r^2 = (6.67 × 10^-11)(4.0)(6.0)/(0.50)^2 = 6.4 × 10^-9 N. The force is attractive and acts along the line between the centers of the spheres. - 2
Earth has a mass of 5.97 × 10^24 kg and a radius of 6.37 × 10^6 m. Calculate the gravitational field strength at Earth's surface.
Gravitational field strength is force per kilogram, so use g = GM/r^2.
The gravitational field strength is g = GM/r^2 = (6.67 × 10^-11)(5.97 × 10^24)/(6.37 × 10^6)^2 = 9.81 N/kg. This is the same as 9.81 m/s^2 for the acceleration of a freely falling object near Earth's surface. - 3
A 70.0 kg astronaut stands on the surface of a planet where the gravitational field strength is 3.7 N/kg. Calculate the astronaut's weight on that planet.
The astronaut's weight is W = mg = (70.0 kg)(3.7 N/kg) = 259 N. The astronaut would weigh about 260 N on that planet. - 4
A satellite is in a circular orbit around Earth at a distance of 7.00 × 10^6 m from Earth's center. Use Earth's mass as 5.97 × 10^24 kg. Calculate the satellite's orbital speed.
Set gravitational force equal to centripetal force to get v = sqrt(GM/r).
For a circular orbit, v = sqrt(GM/r). Substituting gives v = sqrt((6.67 × 10^-11)(5.97 × 10^24)/(7.00 × 10^6)) = 7.55 × 10^3 m/s. The satellite's orbital speed is about 7550 m/s. - 5
Using the satellite in problem 4 with orbital radius 7.00 × 10^6 m and speed 7.55 × 10^3 m/s, calculate the time for one complete orbit in seconds.
The satellite travels one circumference, 2πr, during one orbit.
The orbital period is T = 2πr/v = 2π(7.00 × 10^6)/(7.55 × 10^3) = 5.83 × 10^3 s. One orbit takes about 5830 seconds, or about 97 minutes. - 6
If the distance between two objects doubles while their masses stay the same, how does the gravitational force between them change?
The gravitational force becomes one fourth as large. This happens because gravitational force follows an inverse-square relationship with distance, so doubling the distance gives 1/2^2 = 1/4 of the original force. - 7
A spacecraft moves from a distance r from a planet's center to a distance 3r. How does the gravitational field strength change?
Compare the new distance with the old distance using the inverse-square relationship.
The gravitational field strength becomes one ninth as large. Since g = GM/r^2, increasing the distance to 3r changes the field strength by a factor of 1/3^2 = 1/9. - 8
A moon of mass 7.35 × 10^22 kg orbits a planet of mass 5.97 × 10^24 kg. The distance between their centers is 3.84 × 10^8 m. Calculate the gravitational force between them.
The gravitational force is F = Gm1m2/r^2 = (6.67 × 10^-11)(5.97 × 10^24)(7.35 × 10^22)/(3.84 × 10^8)^2 = 1.99 × 10^20 N. The planet and moon pull on each other with equal and opposite forces of this magnitude. - 9
Explain why an astronaut in the International Space Station feels weightless even though Earth's gravity is still acting on the astronaut.
Weightlessness in orbit is about free fall, not the absence of gravity.
The astronaut feels weightless because the astronaut and the space station are both in continuous free fall around Earth. Gravity provides the centripetal acceleration needed for orbit, so there is no normal force from the floor pushing up on the astronaut. - 10
A planet has twice Earth's mass and the same radius as Earth. Compared with Earth, what is the gravitational field strength at the planet's surface?
The planet's surface gravitational field strength is twice Earth's. Since g = GM/r^2, doubling the mass while keeping the same radius doubles g, so the field strength would be about 19.6 N/kg. - 11
A satellite orbiting Earth moves to a higher circular orbit. Describe how its orbital speed and orbital period change.
For circular orbits around the same planet, larger orbital radius means lower speed and longer period.
In a higher circular orbit, the satellite's orbital speed decreases because v = sqrt(GM/r). Its orbital period increases because the orbit has a larger circumference and the satellite moves more slowly. - 12
A geostationary satellite stays above the same point on Earth's equator. Explain the two main orbital conditions needed for this to happen.
A geostationary satellite must orbit in Earth's equatorial plane and have an orbital period equal to Earth's rotation period, about 24 hours. These conditions allow it to remain above the same point on the equator as Earth rotates.