Energy and gravitation problem

In summary, NASA is considering using solar sailing as a means of propulsion for spacecrafts. The sails would be made of a material with a density of 1000kg.m^{-3}. To balance the gravitational attraction of the sail towards the sun, the minimum thickness of the sail would depend on the distance from the sun and the solar constant, which is 1.4kW.m^{-2} at the Earth's orbit. The force and mass of the sail are both directly proportional to the area of the sail, which can be represented by A. Therefore, the area of the sail will cancel out in the equation of equilibrium, similar to the equation of planetary motion.
  • #1
NewtonianAlch
453
0

Homework Statement


NASA is considering solar sailing: using the momentum of light and of massive particles emitted from the sun to help push a spacecraft equipped with large, diaphanous sails. Assume that the density of the material from which the sails are made is about 1000kg.m[itex]^{-3}[/itex].

a) What is the minimum thickness of a sail such that the use of the momentum from the sun's light is enough to balance the gravitational attraction (of the sail alone) towards the sun? At the Earth's orbit (150 million km), the "solar constant" or intensity of solar radiation is: 1.4kW.m[itex]^{-2}[/itex].

m[itex]_{sun}[/itex] = 1.99E30, G = 6.67E-11

b) How does the answer depend on distance from the sun?


Homework Equations



Force = Gmm/r^2
Volume = Area x Thickness
Density = Mass/Volume

The Attempt at a Solution



I'm not entirely sure how to start this. I guess I would need to find out the Area somehow, as well as the mass of the sail, that could help determine the gravitational attraction.
 
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  • #2
Consider the area 1 m^2.

ehild
 
  • #3
You might also want to investigate "radiation pressure".
 
  • #4
ehild said:
Consider the area 1 m^2.

ehild

Where did you get that from?
 
  • #5
The given data are the density of the sail material and the intensity of the light. So both mass and force are directly proportional to the area of the sail. The area will cancel from the equation of equilibrium, just like it cancels in the equation of planetary motion: it does not matter if a fly orbits around a planet or a big space station, they move with the same velocity at the same orbit. If you do not like 1m2 for the area, just call it A.

ehild
 

Related to Energy and gravitation problem

1. What is energy?

Energy is the ability to do work or cause change. It exists in many forms, such as kinetic energy (energy of motion), potential energy (stored energy), and thermal energy (heat). Energy can be converted from one form to another, but it cannot be created or destroyed.

2. How does gravity affect energy?

Gravity is a fundamental force that causes objects with mass to be attracted to one another. It affects energy by changing the potential energy of an object based on its distance from the center of the gravitational field. For example, an object held high above the ground has more potential energy due to the force of gravity pulling it towards the Earth.

3. What is the relationship between energy and motion?

Energy and motion are closely related. An object in motion has kinetic energy, which is the energy of its movement. In turn, kinetic energy can be converted into other forms of energy, such as thermal energy, when an object comes to a stop due to friction.

4. How does energy relate to the laws of thermodynamics?

The laws of thermodynamics govern the transfer and transformation of energy. The first law states that energy cannot be created or destroyed, only transferred or converted. The second law states that energy tends to disperse and become less organized over time. These laws apply to all forms of energy, including gravitational potential energy.

5. What is the equation for calculating potential energy due to gravity?

The equation for calculating potential energy due to gravity is PE = mgh, where PE is potential energy, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s² near the Earth's surface), and h is the height of the object above the ground. This equation only applies to objects near the surface of the Earth, as the value of g changes with distance from the Earth's center.

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