Weight of object below the Earth's surface

In summary: Please post new content in the forum. Thank you.In summary, the textbook is wrong if the given radius is correct.
  • #1
rrosa522
48
0

Homework Statement



Calculate the gravitational force acting on a 9.6-kg mass...?
1.5 x 10^4 m above Earth's center. How do i solve this??

Homework Equations

The Attempt at a Solution


the answer is 17N but i don't understand how
 
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  • #2
Hi rrosa522. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

What is the equation for gravitational force between m1 and m2 spaced a distance r apart? If not in your textbook, use google.
 
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  • #3
This chap Isaac Newton had stuff like this figured out.
 
  • #4
SteamKing said:
This chap Isaac Newton had stuff like this figured out.
i must admit that after reading this threads title i was disappointed by its content XD
 
  • #5
try and use this equation:
[itex] F_{grav} = \frac{Gm_{1}m_{2}}{r^2} [/itex]
Where m1 and m2 are two masses, r is a distance and G is a constant
 
  • #6
_N3WTON_ said:
try and use this equation:
[itex] F_{grav} = \frac{Gm_{1}m_{2}}{r^2} [/itex]
Where m1 and m2 are two masses, r is a distance and G is a constant
i tried the formula but i don't get 17N
 
  • #7
rrosa522 said:
i tried the formula but i don't get 17N
you should have the mass of Earth (its probably in your textbook) we can call this m1, the other mass is given to you. You have the distance of the object, so enter these values into the formula given above
 
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  • #8
Better yet, show me what you are entering in the formula
 
  • #9
A distance of 1.5 x 10^4 m from the Earth's center is well below the Earth's surface, and within the inner core. How did you apply Newton's gravitational force law for that location? Show your work.

Where did your answer of 17N come from ?

Also, your thread title does not meet Physics Forums requirements; thread titles should be descriptive of the thread content, describing the problem or area of physics involved. Please note this for future questions that you post. I have changed the title to be more appropriate.
 
  • #10
I'm sorry, I read the problem as 1.5x10^4 from the Earth's surface...either way show how you are using the formula
 
  • #11
rrosa522 said:
i tried the formula but i don't get 17N
_N3WTON_ said:
I'm sorry, I read the problem as 1.5x10^4 from the Earth's surface...either way show how you are using the formula
(6.67x10^-11) (9.6) (5.97 x 10^24) /
(1.5 x 10^4)^2
the answer i get 1.70x10^7 N
(sorry i didnt add units in my solution, its just gets confusing for me)
the answer at the the back of my textbook is 17 N, i don't understand why
 
  • #12
gneill said:
A distance of 1.5 x 10^4 m from the Earth's center is well below the Earth's surface, and within the inner core. How did you apply Newton's gravitational force law for that location? Show your work.

Where did your answer of 17N come from ?

Also, your thread title does not meet Physics Forums requirements; thread titles should be descriptive of the thread content, describing the problem or area of physics involved. Please note this for future questions that you post. I have changed the title to be more appropriate.
(6.67x10^-11) (9.6) (5.97 x 10^24) /
(1.5 x 10^4)^2
the answer i get 1.70x10^7 N
the answer at the the back of my textbook is 17 N, i don't understand why
 
  • #13
The textbook is wrong if the radius that you specified is correct.

17 N would be correct if the given radius was in km not meters. That would place the 9.6 kg object well above the Earth's surface. Check the units on the given radius.
 
  • #14
let take one object as Earth and other as the object
there distance as 1.5 x 10^4
f=gM1m2/d^2
 
  • #15
you are using meters and kilometres. This is not giving correct answer due to this reason. Use either kilometres or metres only. to make the switch easy go from (1.5x10^4KM) to (1500x10^4M)
 
  • #16
physicsisfunfightme said:
you are using meters and kilometres. This is not giving correct answer due to this reason. Use either kilometres or metres only. to make the switch easy go from (1.5x10^4KM) to (1500x10^4M)
Hello, @physicsisfunfightme .

:welcome:

The thread you replied to is more than 7 years old.
 

FAQ: Weight of object below the Earth's surface

What is the weight of an object below the Earth's surface?

The weight of an object below the Earth's surface is determined by its mass and the strength of the gravitational pull at that location.

Does the weight of an object change below the Earth's surface?

Yes, the weight of an object changes below the Earth's surface due to the varying strength of gravity at different depths.

How does depth affect the weight of an object below the Earth's surface?

The deeper an object is below the Earth's surface, the stronger the gravitational pull towards the center of the Earth, resulting in an increase in weight.

Can the weight of an object below the Earth's surface be negative?

No, the weight of an object below the Earth's surface cannot be negative as weight is a measure of the force of gravity on an object and cannot be less than zero.

What is the unit of measurement for weight below the Earth's surface?

The unit of measurement for weight below the Earth's surface is the same as on the surface, which is typically expressed in Newtons (N) or pounds (lb).

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