Gravitation attraction of a mountain

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The discussion focuses on calculating the mass of a mountain using gravitational attraction principles. A user attempted to derive the mountain's mass based on the gravitational force acting on a ball, but others pointed out flaws in their approach, particularly the need for a free body diagram to accurately represent forces. The importance of including tension in the string and recognizing that the ball is motionless at a slight angle is emphasized. Participants stress that the equations presented are incomplete without incorporating the mountain's mass and proper vector analysis. The thread concludes with requests for clearer diagrams and more detailed calculations to facilitate further assistance.
ac7597
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Homework Statement
Mason and Dixon are surveying a peculiar region: a flat plain with a single giant mountain sticking up out of the ground. The mountain has a very round shape. "It looks like the mountain is sitting on the plain," says Dixon, "but miners have discovered that the mountain extends down below the surface. It appears that the mountain is half-buried in the dirt."

When they are still a distance 15000 meters away from the mountain, they make a local gravitational force measurement: they hang a ball of mass 11.5 kg from a string. The string doesn't hang vertically, but tilts very slightly towards the mountain. Dixon estimates the angle to be 0.00050 degrees.

Estimate the mass of the mountain.

The mountain is shaped like a giant sphere (half of which is buried underground), and appears to be made of ordinary rock. Estimate its height above the ground.
Relevant Equations
F=G(Mass)(mass)/(radius)^2
G=6.72 * 10^(-11)
I tried to solve for mass of the mountain by:

(mass of ball) (9.8m/s^2)= G(mass of ball)(mass of mountain)/ (15000m)^2
The mass of the ball cancels out leaving with mass of mountain=33.04 * 10^(18) kg.
 
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Your equation says that the force with which the Earth attracts the ball is equal to the force with which the mountain attracts the ball. That is simply not true. Draw a free body diagram showing all the forces acting on the ball and find the resultant by vector addition. You know that the resultant should be at 0.00050 degrees.
 
y direction: (11.5kg)(9.8m/s^2)sin(0.0005) =0.983 * 10^(-3) N
x direction: ((11.5kg)(9.8m/s^2)cos(0.0005) =112.7 N ?
 
ac7597 said:
y direction: (11.5kg)(9.8m/s^2)sin(0.0005) =0.983 * 10^(-3) N
x direction: ((11.5kg)(9.8m/s^2)cos(0.0005) =112.7 N ?
Directions x and y are meaningless without a diagram. Please show and post a diagram showing all the forces acting on the ball and label them appropriately. Write equations derived from that diagram. Don't forget that you need to introduce the unknown mass of the mountain. The equations in post #3 are independent of it.
 
kuruman said:
Directions x and y are meaningless without a diagram. Please show and post a diagram showing all the forces acting on the ball and label them appropriately. Write equations derived from that diagram. Don't forget that you need to introduce the unknown mass of the mountain. The equations in post #3 are independent of it.
what he said (very small).jpg
 
I got up to m(ax) =m1(mx)(G)/(r^2)-m1(g)cos(0.0005)
 

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Is the ball suspended in mid air with only the two gravitational forces acting on it?
 
The question doesn't specify but I think when it said 'hang vertically' it means it's in midair. I can't solve for (mx) without knowing the acceleration (ax).
 
ac7597 said:
When they are still a distance 15000 meters away from the mountain, they make a local gravitational force measurement: they hang a ball of mass 11.5 kg from a string. The string doesn't hang vertically, but tilts very slightly towards the mountain. Dixon estimates the angle to be 0.00050 degrees.
What about the string?
 
  • #10
assume string is massless
 
  • #11
ac7597 said:
assume string is massless
That's not what @kuruman was asking. Isn't the string attached to the mass? Is the string slack?
 
  • #12
String is not slack. Is the body diagram correct?
 
  • #13
ac7597 said:
String is not slack. Is the body diagram correct?
The free body diagram is not correct. If the string is not slack, it follows that it must be under tension. This tension acts on the ball. Add it to your diagram. Also, the acceleration of the ball is zero; the ball just hangs there motionless at the given very small angle relative to the vertical. Should you require more help, please post your revised diagram and a complete revised attempt at a solution.
 
  • #14
kuruman said:
The free body diagram is not correct. If the string is not slack, it follows that it must be under tension. This tension acts on the ball. Add it to your diagram. Also, the acceleration of the ball is zero; the ball just hangs there motionless at the given very small angle relative to the vertical. Should you require more help, please post your revised diagram and a complete revised attempt at a solution.
AND, please make it larger / more legible this time.
 
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  • #15
Is this the answer for part 1?
 

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  • #16
ac7597 said:
Is this the answer for part 1?
Please re-read post #13 and post #14
 
Last edited:
  • #17
ac7597 said:
Is this the answer for part 1?
The thumbnail sketch in the middle looks right, but it should be several times as large.
You don't need to show the mountain in an FBD of the bob, just the force it exerts.
None of the rest belongs in an image. Per forum rules, please take the trouble to type in your working. Not only does that make it much easier to read, it is also easier to comment on specific steps.
 

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