Weight of a person a certain distance above the Earth

[AfronPie]

Homework Statement

The problem is stated in the picture.

Homework Equations

Fg=Gm.1m.2/r^2

The Attempt at a Solution

I found a similar question where someone wanted to find how high will the weight be half the weight on the surface. Here is one of the answers:
F =G.m1.m2/r^2
F´ =G.m1.m2/r´^2
so F/F´= r´^2/r^2 = (r´/r)^2.
So I tried applying it to my situation. I wrote 34/100=r'^2/(6.38*10^6)^2. Solving for r' I got an answer of x=3.72*10^6meters=3720km. The correct answer of 4600km is displayed in the screen shot attached. I really don't understand how to do these problems, though I do understand that Fg is inverse square proportional to r. Any help would be greatly appreciated. Thanks.

 

[scottdave]

Look again at your F and F'. You have them reversed. Then you will be solving for radius from center of Earth. You need to find height above the surface.

 

[AfronPie]

Look again at your F and F'. You have them reversed. Thwn you will be solving for radius from center of Earth. You need to find height above the surface.

Can you please elaborate. How would I rewrite my original equation of 34/100=r'^2/(6.38*10^6)^2 .?

 

[ElectricRay]

Am I mistaking something here? I thought we could write the equation as follows.

$$F = \frac {G*M1*M2} {r^2}$$

$$\frac F {F'} = \frac {G*M1*M2} {r^2} * \frac {r'^2} {G*M1*M2} = \frac {r'^2} {r^2}$$

Maybe I make a huge mistake so if I confuse the OP I am sorry

 

[ElectricRay]

Ahhh ok @scottdave ur right I see it my apologizes

F and F' are reversed in OP formula

 

[dean barry]

is it not easier to disregard m2 ? (persons mass) the solution will be all but identical

 

[ElectricRay]

Well in fact that's what's happening as far I understand. You disregard M1, M2 and G the only thing that matters here is the ratio of you weight with respect to when your on the surface of the Earth and on a distance such that your weight is 34% less. For that you don't need M1, M2 and G because your weight is proportional to the inverse of r^2. I hope I'm using the right words here.

 

[AfronPie]

Am I mistaking something here? I thought we could write the equation as follows.

$$F = \frac {G*M1*M2} {r^2}$$

$$\frac F {F'} = \frac {G*M1*M2} {r^2} * \frac {r'^2} {G*M1*M2} = \frac {r'^2} {r^2}$$

Maybe I make a huge mistake so if I confuse the OP I am sorry

Using your formula, I am still confused. I wrote the equation as F/F'=100/34=r'^2/(6.38*10^6)^2. Solving this for r' I get 10,940 km. I really don't get what I plugged in wrong. F/F' indicated that the original force is around three times greater then the ' force, and r is the radius of the earth.

 

[ElectricRay]

Have you swopped F and F' ?

 

[ElectricRay]

As far I understand it should be like this:

$$\frac {F'} F = \frac {G*M1*M2} {r'^2} * \frac {r^2} {G*M1*M2} = \frac {r^2} {r'^2}$$

Now it is important to think as that in the question is written that your weight reduces with 34%, so you can't plug in 34% you need to take care of that. If I do that I come on 4520 km. I think that's correct.

 

[AfronPie]

As far I understand it should be like this:

$$\frac {F'} F = \frac {G*M1*M2} {r'^2} * \frac {r^2} {G*M1*M2} = \frac {r^2} {r'^2}$$

Now it is important to think as that in the question is written that your weight reduces with 34%, so you can't plug in 34% you need to take care of that. If I do that I come on 4520 km. I think that's correct.

How do you incorporate 34% into the equation without plugging in 34/100 or 100/34 to the equation?

 

[ElectricRay]

The question states you weight is reduced with 34% so what percentage is left?

 

[AfronPie]

The question states you weight is reduced with 34% so what percentage is left?

The question states you weight is reduced with 34% so what percentage is left?

You have 66 percent of your weight left. But even if I plug in 66 I get: 66/100=(6.38*10^6)^2/r'^2, getting an r' of 7853km?

 

[ElectricRay]

$$0.34 = \frac {6380^2} {r'^2}$$

Then you get the distance from the centre of the earth, but they ask the distance form the surface of the earth. I am sorry if i don't explain myself good enough.

I come on 4561 km

 

[AfronPie]

$$0.34 = \frac {6380^2} {r'^2}$$

Then you get the distance from the centre of the earth, but they ask the distance form the surface of the earth. I am sorry if i don't explain myself good enough.

I come on 4561 km

Thanks, I made 2 mistakes. I switched F and F', and also I didn't know that you had to subtract the distance 6371km but now I understand. Thanks for your help.

 

[ElectricRay]

The key is in this problem is knowing that this formula applies from the centre of the mass and the question that was asked was from the surface of the mass.

Well if you swop everything you still come on the same answer:
$${r'}=\sqrt{ {6380^2} * \frac 1 {0.34}} = 10941.6$$
$$10941 - 6380 = 4561.6 km$$

I think it is a way of interpreting and applying the formulas I think. Thank you too because with trying to help you I learned something too ;)

 

[ElectricRay]

I didn't know that you had to subtract the distance 6371km but now I understand. Thanks for your help.

Yes we got a small differnece I used your 6380 km but in fact according google it is indeed 6371 km

 

[epenguin]

As you know of course if you weighed yourself up there using those Medic's balances where they load weights on some counterbalance you would measure yourself as the same weight as here. But if you weighed yourself with a spring balance that had been calibrated at the Earth's surface you would get your result.

 

[scottdave]

As a bonus, can you figure how the astronauts on the International Space Station (altitude approx 320 km) experience a weightless environment?

 

[ElectricRay]

As you know of course if you weighed yourself up there using those Medic's balances where they load weights on some counterbalance you would measure yourself as the same weight as here. But if you weighed yourself with a spring balance that had been calibrated at the Earth's surface you would get your result.

Yes because everything het 34% less right? Please don't tell me the calculation is still wrong ;)

 

[ElectricRay]

As a bonus, can you figure how the astronauts on the International Space Station (altitude approx 320 km) experience a weightless environment?

This is (please correct me if I am wrong) because they are continuously falling to the centre of the earth. But they have such amount of speed that they keep circling arpund earth.
Would ISS have less speed it wpuld be catched and pulled to earth. This weightless environment is like what you feel when your in an elevator that goes fast down.

 

[scottdave]

This is (please correct me if I am wrong) because they are continuously falling to the centre of the earth. But they have such amount of speed that they keep circling arpund earth.
Would ISS have less speed it wpuld be catched and pulled to earth. This weightless environment is like what you feel when your in an elevator that goes fast down.

Yes, that's it. I only learned this about 2 years ago.

 

[ElectricRay]

I never thought about it and after the initial question we found a solution but still wasn't thinking about ISS. SO then you came with the bonus and i started to thinking. Just by applying the laws as far as i know it I could give this answer. That was cool! But it al makes sense. People think in space there is no gravity but that is wrong plug in some numbers in the original formula we used and you'll see it never gets to 0.