How Do You Calculate Changes in Gravitational Force and Weight?

In summary: That's correct. Remember, the point is to check your answer. If you get 1.56 m/s^2 and they wanted the distance in meters, then you definitely did something wrong.
  • #1
Neek 007
41
0

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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  • #2
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?
 
  • #3
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
 
  • #4
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
 
  • #5
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.
 
  • #6
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
 
  • #7
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
 
  • #8
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?
 
  • #9
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
 

Related to How Do You Calculate Changes in Gravitational Force and Weight?

1. What is circular motion?

Circular motion is a type of motion in which an object moves in a circular path around a central point or axis. This type of motion is characterized by a constant speed and a constantly changing direction.

2. How is circular motion related to gravitation?

Circular motion is related to gravitation through the force of gravity. According to Newton's Law of Universal Gravitation, any two objects with mass will experience an attractive force between them. In circular motion, the centripetal force that keeps an object moving in a circular path is provided by the force of gravity.

3. What is centripetal force?

Centripetal force is the force that keeps an object moving in a circular path. It is always directed towards the center of the circle and is necessary to maintain the object's circular motion. In circular motion, this force is provided by gravity, tension, or any other force that acts towards the center of the circle.

4. What is the relationship between circular motion and velocity?

Circular motion and velocity are closely related. The velocity of an object in circular motion is constantly changing because its direction is constantly changing. However, the speed of the object remains constant. This is because velocity takes into account both speed and direction, and in circular motion, only the direction changes, not the speed.

5. How is circular motion different from linear motion?

Circular motion and linear motion are fundamentally different because they involve different types of motion. Linear motion is the motion of an object in a straight line with a constant speed, while circular motion involves an object moving in a circular path with a constant speed and a constantly changing direction. Additionally, linear motion is described using distance and time, while circular motion is described using angular displacement and time.

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