Gravitational acceleration, cosine problem

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SUMMARY

The gravitational acceleration at latitude x can be modeled by the equation g(x) = a*cos(2x) + b. To determine the constants a and b, the gravitational acceleration values at x=0 (9.780 m/s²) and x=90 (9.832 m/s²) were used. The calculations revealed that the cosine function must be evaluated in degrees, correcting the initial misunderstanding that radians were applicable. Ultimately, the correct values for a and b can be derived from the equations 9.780 = a + b and 9.832 = -1*a + b.

PREREQUISITES
  • Understanding of trigonometric functions, specifically cosine.
  • Knowledge of gravitational acceleration concepts.
  • Ability to solve linear equations.
  • Familiarity with degrees versus radians in trigonometry.
NEXT STEPS
  • Learn how to solve systems of linear equations using substitution and elimination methods.
  • Study the properties of the cosine function and its applications in physics.
  • Explore the differences between degrees and radians in trigonometric calculations.
  • Investigate real-world applications of gravitational acceleration in different latitudes.
USEFUL FOR

Students studying physics, mathematics, or engineering, particularly those focusing on gravitational forces and trigonometric applications.

BadatPhysicsguy
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Homework Statement


The gravitational acceleration at latitude x (0<x<90) can be estimated with g(x)=a*cos(2x)+b.
1) Determine what a and b is if the gravitational acceleration is 9.780m/s^2 at x=0 and 9.832m/s^2 at x=90.

Homework Equations

The Attempt at a Solution


So I begin by entering what I know:
9.780=a+b (Because cos(2*0) is 1)
9.780-a=b
And then the other equation:
9.832=-5.9846*a+b (Because cos(2*90) is -5.9846)

I then combine the two equations, jumping over the middle steps, gives:
-0.00868=a-a which means... -0.00868=0 which isn't logical. I wanted to get what a is, then enter it into either of the equations and get b.
 
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BadatPhysicsguy said:

Homework Statement


The gravitational acceleration at latitude x (0<x<90) can be estimated with g(x)=a*cos(2x)+b.
1) Determine what a and b is if the gravitational acceleration is 9.780m/s^2 at x=0 and 9.832m/s^2 at x=90.

Homework Equations

The Attempt at a Solution


So I begin by entering what I know:
9.780=a+b (Because cos(2*0) is 1)
9.780-a=b
And then the other equation:
9.832=-5.9846*a+b (Because cos(2*90) is -5.9846)

cos(2*90) = -5.9846? cosine can only evaluate between -1 and +1. Remember, latitude is measured in degrees, not radians. Try again.
 
BadatPhysicsguy said:

Homework Statement


The gravitational acceleration at latitude x (0<x<90) can be estimated with g(x)=a*cos(2x)+b.
1) Determine what a and b is if the gravitational acceleration is 9.780m/s^2 at x=0 and 9.832m/s^2 at x=90.

Homework Equations

The Attempt at a Solution


So I begin by entering what I know:
9.780=a+b (Because cos(2*0) is 1)
9.780-a=b
And then the other equation:
9.832=-5.9846*a+b (Because cos(2*90) is -5.9846)
You're skipping some steps here and writing stuff that isn't true. cos(2 * 90°) = -1.
BadatPhysicsguy said:
I then combine the two equations, jumping over the middle steps, gives:
-0.00868=a-a which means... -0.00868=0 which isn't logical. I wanted to get what a is, then enter it into either of the equations and get b.
 
SteamKing said:
cos(2*90) = -5.9846? Remember, latitude is measured in degrees, not radians. Try again.
Thank you! The book said that real life applications are measured in radians and not degrees, so I assumed it would be the same here. I solved it now, thanks again!
 
BadatPhysicsguy said:
Thank you! The book said that real life applications are measured in radians and not degrees, so I assumed it would be the same here. I solved it now, thanks again!

Your book is not correct on this point. Latitude and longitude are always measured in degrees. In most other applications, degrees or grads are used. Radians are generally used only in math or science, because they simplify working with derivatives and integrals of trig functions.
 

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