Ratio of Distances using Acceleration of Gravity

In summary, the problem involves a constant-density planet of radius R with a gravitational acceleration of ag = as at its surface. There are two distances from the center of the planet at which ag = as/2. To find the ratio of these distances, the expressions for ag must be written for each distance, taking into account the symmetries of the problem. This leads to the conclusion that the radius and density distribution uniquely determine the mass of the planet. The correct approach for finding r1 is to set ar1 = as/2, while for r2, the expression should be ar2 = as.
  • #1
Kabal
2
0

Homework Statement



A constant-density planet of radius R has a gravitational acceleration ag =
as at its surface. There are two distances from the center of the planet at
which ag = as/2. Show that the ratio of these distances may be given by
R = 2√2


Homework Equations



ag= GM/R2

ρ=M/V


The Attempt at a Solution



This is my attempt at approaching the problem. Given that there are two distances I'm thinking that one of the distances is inside the planet while the other would be outside. I know that this is a constant density planet therefore in some fashion I must use a density equation. The problem I'm receiving though is that when I setup the ratio I simply get 1

GM/R2 = GρV/2r2

GM/R2 = Gρ(4/3)∏r13/2r12

2M/R2 = ρ(4/3)∏r1

3M/2∏R2ρ=r1

At this point I simply do the same thing for r2 then set the ratio.
This definitely doesn't feel right and I think I'm missing a key understanding to be able to completely understand this problem completely. Another idea I had was that the Rs must vary depending on where in relation I'm talking about. Could it be that the radii is R-r1 and R+r2?

Any help is much appreciated.
 
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  • #2
Think about where the two distances at which ##a_g = a_s / 2## must be, in general terms, taking into account the symmetries of the problem. Then write down the expressions for ##a_g## that apply to each distance.

Your last sentence shows that you are on the right track.
 
  • #3
Thanks I'll give that an attempt

Another question. Is it possible that the planet with radius R have different masses?
Ex.
GM/2R2=Gm/R2

I feel that this is wrong because of the fact that the planet is at constant density. Therefore at Radius R it must have mass M.
 
  • #4
The radius and density distribution uniquely determine the mass. Ignoring such real-world things as infalling mass, of course. In other words, since in the problem the density is constant and the radius is fixed, there is only one possible mass for the planet, or for any planet with those two characteristics. But it isn't the constancy of the density per se that makes that true; any density distribution would also give a single mass, as long as it is the same for the planets under consideration.
 
  • #5
What you did for r1 was correct but you can't do the same thing for r2 since r2 is not inside the planet.
 
  • #6
Kabal said:
GM/R2 = GρV/2r2
No, this is backwards. as = GM/R2, and you want an r1 for which ar1 = as/2.
All that is for r1 < R. Please post your working for the r2 > R case.
 

Related to Ratio of Distances using Acceleration of Gravity

What is the ratio of distances using acceleration of gravity?

The ratio of distances using acceleration of gravity is a mathematical formula used to calculate the distance an object travels due to the force of gravity. It is expressed as d1/d2 = (t1/t2)^2, where d1 and d2 are the distances traveled and t1 and t2 are the times taken for an object to travel those distances. This ratio is commonly used in physics and engineering to understand the motion of objects under the influence of gravity.

How is the ratio of distances using acceleration of gravity related to the acceleration of gravity?

The acceleration of gravity, denoted as 'g', is a constant value that represents the rate at which an object accelerates towards the Earth due to gravity. The ratio of distances formula includes this value, as it depends on the acceleration of gravity to calculate the distance traveled. This means that the value of g will affect the final result of the ratio of distances calculation.

What is the significance of the ratio of distances using acceleration of gravity in physics?

The ratio of distances using acceleration of gravity is significant in physics as it helps us understand the motion of objects under the influence of gravity. By using this formula, we can calculate the distance an object will travel in a given amount of time, or the time it will take for an object to travel a certain distance. This is essential in fields such as mechanics, astronomy, and engineering.

How is the ratio of distances using acceleration of gravity different from the formula for free fall?

The formula for free fall, d = 1/2gt^2, only calculates the distance an object travels when it is dropped from a certain height. It does not take into account the distance traveled at different time intervals. On the other hand, the ratio of distances formula takes into account both the distances and times at two different intervals and compares them. This makes it a more comprehensive formula for understanding the motion of objects under gravity.

Can the ratio of distances using acceleration of gravity be used in real-world scenarios?

Yes, the ratio of distances using acceleration of gravity can be used in real-world scenarios. For example, it can be used to calculate the distance a ball will travel if thrown at a certain angle and speed, or the time it will take for a rollercoaster to reach the ground from the top of a hill. This formula is applicable in any scenario where an object is under the influence of gravity, and its motion can be measured.

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