Potential energy of what? Is this equation relevant for the problem?Raghav Gupta said:Homework Statement
Assume that the orbital radius of the Earth decreases from R to R-ΔR due to air drag, provided the change is very small compared to the radius of earth.
Then what is change in orbital velocity, change in kinetic energy, change in potential energy and work done by air resistance?
Homework Equations
F = Gm1m2/r2
Kinetic energy = 1/2 mv2
Potential energy = mgh
How do you get the work of a force? Is Gm1m2/r2 the air resistance?Raghav Gupta said:The Attempt at a Solution
Work done by air resistance is Gm1m2Δr/r2 ?
BvU said:My impression (more like a proven hypothesis) was always that central forces conserve angular momentum and I don't see anything non-central in this.
Well I was looking potential energy in Wikipedia and they say it is because of some field.ehild said:Potential energy of what? Is this equation relevant for the problem?
How do you get the work of a force? Is Gm1m2/r2 the air resistance?
Raghav Gupta said:Well I was looking potential energy in Wikipedia and they say it is because of some field.
They have written for capacitor ½ CV2 is potential energy. Why it is called potential energy! Because it is of electric field?
Whenever someone calls kinetic energy ,the first thing comes to my mind is ½mv2.
But I know it is energy due to motion.
The work expression is wrong for air resistance, I know. I was not able to imagine it properly.
Work done = Force * displacement cosθ
You can assume circular orbit and assume that the one with smaller radius is also a stable orbit. Knowing the radius, you get the speed with comparing the centripetal force to the force of gravity.Raghav Gupta said:I think one can use dimensional analysis to match options only helping for orbital velocity.
But what is the correct logical way?
Presumably this is meant to be the orbital radius of an Earth-orbiting satellite, yes?Raghav Gupta said:Assume that the orbital radius of the Earth decreases from R to R-ΔR due to air drag, provided the change is very small compared to the radius of earth.
Yes I think so. As Earth is so big and air drag on it does not make sense.gneill said:Presumably this is meant to be the orbital radius of an Earth-orbiting satellite, yes?
Gravitation is a fundamental force of nature that describes the attraction between objects with mass. It is responsible for the motion of planets, stars, galaxies, and other celestial bodies.
Gravitation affects everyday life in several ways. It keeps us grounded on Earth, causes objects to fall to the ground, and keeps planets in orbit around the sun. It also plays a role in tides, ocean currents, and the stability of structures such as buildings and bridges.
Mass is a measure of the amount of matter an object contains, while weight is a measure of the force of gravity on an object. Gravitation affects weight, but not mass. For example, an object will have the same mass on Earth and on the moon, but it will weigh less on the moon due to the moon's weaker gravitational pull.
The force of gravitation between two objects decreases as the distance between them increases. This is described by the inverse square law, which states that the force is inversely proportional to the square of the distance between the objects. This means that the force decreases rapidly as distance increases.
The gravitational constant (G) is a fundamental constant in physics that determines the strength of the gravitational force between two objects. Its value is used in calculations involving gravitation and is essential for understanding the behavior of celestial bodies in our universe.