How Do You Calculate Changes in Gravitational Force and Weight?

AI Thread Summary
To determine how far above Earth's surface a person's weight is halved, the gravitational acceleration must be reduced to 4.9 m/s². The correct approach involves using the gravitational formula and solving for height above the surface, ensuring to account for Earth's radius. For the second problem, the gravitational attraction between a 4 kg and a 2 kg ball positioned 40 cm apart requires converting the distance to meters, as incorrect units led to calculation errors. The discussions emphasize the importance of unit consistency and the correct application of gravitational equations to solve these problems accurately.
Neek 007
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Homework Statement


(These are problems I got wrong on a test)
1. How far above the Earth's surface will a person's weight be reduced to half its value at the surface?

2. A 4kg ball and a 2 kg ball are positioned so that their centers are 40 cm apart. With what force do they attract each other?


Homework Equations


1. rE=6380km=6.38*106m
me=5.98*1024kg
g=(Gm)/r2
G=6.67*10-11Nm2/kg2 where G is universal gravitation


2. G=6.67*10-11Nm2/kg2 where G is universal gravitation
F=(Gm1m2)/r2


The Attempt at a Solution


1. In problem 1 I know that in order for a mass to be half of what it is on Earth's surface, g must be half, so g=9.8m/s2 on Earth's surface, therefore g=4.9m/s2 at r.
What I did was use for formula g=(Gm)/r2 and solve for r. I think for m though, i used 25. For what reason, i don't know. I got the question wrong, so that's not the right method.

2. For problem 2, I used the F=(Gm1m2)/r2 and solved for F, the force. I ended up getting 1.339*10-12. I thought I did this right, but obviously not.

I need some direction into where to correctly start.
 
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Neek 007 said:

Homework Statement


(These are problems I got wrong on a test)
1. How far above the Earth's surface will a person's weight be reduced to half its value at the surface?

2. A 4kg ball and a 2 kg ball are positioned so that their centers are 40 cm apart. With what force do they attract each other?


Homework Equations


1. rE=6380km=6.38*106m
me=5.98*1024kg
g=(Gm)/r2
G=6.67*10-11Nm2/kg2 where G is universal gravitation


2. G=6.67*10-11Nm2/kg2 where G is universal gravitation
F=(Gm1m2)/r2


The Attempt at a Solution


1. In problem 1 I know that in order for a mass to be half of what it is on Earth's surface, g must be half, so g=9.8m/s2 on Earth's surface, therefore g=4.9m/s2 at r.
What I did was use for formula g=(Gm)/r2 and solve for r. I think for m though, i used 25. For what reason, i don't know. I got the question wrong, so that's not the right method.

2. For problem 2, I used the F=(Gm1m2)/r2 and solved for F, the force. I ended up getting 1.339*10-12. I thought I did this right, but obviously not.

I need some direction into where to correctly start.

For problem 1:

In that equation, M is the mass of the Earth.

For problem 2:

You're method is fine. You must have made a calculation error. What numbers did you plug in for r, m1, and m2?
 
G01 said:
For problem 1:

In that equation, M is the mass of the Earth.

Also, they say they want the height above the Earth's surface, not the radius.

You might have approached the problem in terms of ratios. Lot's of things cancel when you do these things as ratios!

g = G*M/Re^2

g/2 = G*M/(Re + h)^2

take the ratio (g/2)/g = 1/2 = ...
solve for h
 
G01 said:
For problem 1:

In that equation, M is the mass of the Earth.

For problem 2:

You're method is fine. You must have made a calculation error. What numbers did you plug in for r, m1, and m2?

for r i used 20. m1=2 m2=4
unless the radius is really 40, then i don't see where i messed up.
i got 1.334x10-12N
 
Neek 007 said:
for r i used 20. m1=2 m2=4
unless the radius is really 40, then i don't see where i messed up.
i got 1.334x10-12N

Units. Always check your units. The separation is specified in cm. 40cm is not the same as 40 meters.
 
G01 said:
For problem 1:

In that equation, M is the mass of the Earth.


I think i have the right answer now, of
9*106 by √(GmE)/g
g being 4.9 m/s2
 
gneill said:
Units. Always check your units. The separation is specified in cm. 40cm is not the same as 40 meters.

there is my problem. I should have seen that.
i should have gotten 1.334*10^-6
 
Neek 007 said:
I think i have the right answer now, of
9*106 by √(GmE)/g
g being 4.9 m/s2

Is that the radial distance or the distance above the surface of the Earth?
They asked for the latter, no?
 
i think for number 2 the mistake was r should be changed to meters
 
  • #10
gneill said:
Is that the radial distance or the distance above the surface of the Earth?
They asked for the latter, no?

that is the distance above Earth's surface, without the radius of Earth added on.
 
  • #11
So let's check your result.

a = G*Me/(Re + 9 x 10^6m)^2

= (6.67 x 10^-11)*(5.98 x 10^24)/(6380 x 10^3 + 9 x 10^6)^2

= 1.56 m/s^2
 
  • #12
gneill said:
So let's check your result.

a = G*Me/(Re + 9 x 10^6m)^2

= (6.67 x 10^-11)*(5.98 x 10^24)/(6380 x 10^3 + 9 x 10^6)^2

= 1.56 m/s^2

how does the acceleration tell me if my answer is right or wrong?
 
  • #13
Neek 007 said:
how does the acceleration tell me if my answer is right or wrong?

oh i see, its the formula for gravity. it should equal 4.9 ms/^2
 
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