Dimensional Analysis Problem: Mass of the Solar System Excluding Planets and Sun

In summary, assuming one hydrogen atom per cubic centimeter in interplanetary space within the solar system and a spherical shape with a radius of 4.0 light-hours, the mass of the solar system excluding discrete bodies such as planets and the sun can be calculated using the speed of light (3.00 x 10^8 m/s) and the number of seconds in a year (3.15 x 10^7). However, without an attempted solution, it is difficult to provide further assistance.
  • #1
HSuzuki
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Homework Statement


Assume that in interplanetary space within the solar system there is one hydrogen atom per cubic centimeter, and that the mass of such an atom is 1.67 x 10^-27 kg. In addition, take the radius of the solar system to be 4.0 light-hours. For the sake of this calculation, take the solar system to be spherical in shape. Note that we are not including the masses of the planets and the sun. Based on this set of assumptions, what is the mass of the solar system excluding discrete bodies such as the planets and the sun? Use for the speed of light 3.00 x 10^8 m/s and assume there are 3.15 x 10^7 seconds in a year.

Thanks for your attempt and help!


Homework Equations





The Attempt at a Solution


 
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  • #2
HSuzuki said:

Homework Statement


Assume that in interplanetary space within the solar system there is one hydrogen atom per cubic centimeter, and that the mass of such an atom is 1.67 x 10^-27 kg. In addition, take the radius of the solar system to be 4.0 light-hours. For the sake of this calculation, take the solar system to be spherical in shape. Note that we are not including the masses of the planets and the sun. Based on this set of assumptions, what is the mass of the solar system excluding discrete bodies such as the planets and the sun? Use for the speed of light 3.00 x 10^8 m/s and assume there are 3.15 x 10^7 seconds in a year.

Thanks for your attempt and help!

Have you tried to solve it? An attempted solution would be nice to see so that we can help you where you need it.
 

FAQ: Dimensional Analysis Problem: Mass of the Solar System Excluding Planets and Sun

What is Dimensional Analysis?

Dimensional analysis is a mathematical method used to convert units from one system to another. It involves using conversion factors to cancel out unwanted units and determine the desired unit.

Why is Dimensional Analysis important in science?

Dimensional analysis is important in science because it helps to ensure accuracy and consistency in measurements. It also allows for easy unit conversions, which are often necessary in scientific calculations.

How do you set up a Dimensional Analysis problem?

To set up a Dimensional Analysis problem, you must first identify the desired unit and the given unit. Next, list out all the conversion factors needed to cancel out the given unit and obtain the desired unit. Finally, multiply all the conversion factors together, making sure to cancel out any common units.

Can Dimensional Analysis be used for any type of unit conversion?

Yes, Dimensional Analysis can be used for any type of unit conversion as long as the given and desired units are related by a conversion factor. It is a versatile method that can be applied to various units of measurement.

What are some common mistakes to avoid when using Dimensional Analysis?

Some common mistakes to avoid when using Dimensional Analysis include using incorrect conversion factors, not canceling out common units, and forgetting to include units in the final answer. It is important to double check all conversions and units to ensure accuracy.

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