Newtons law of universal gravitation question (2 Viewers)

[andrew3]

Would anybody be able to see the answer they get for the following question, i get a different one to the answers(= 1.93 x 10^8 m). I'm not not sure if I'm doing something wrong or its just a wrong answer
Qu) Calculate how far an astronaut would need to be away above the Earth in order for his weight to be 0.01 of his weight on the Earth's surface.
I did the question and got 5.75 x 10^7. Any help would be greatly appreciated, thanks

 

[deswa1]

I just did it quickly and got the same answer as you. Maybe someone else can confirm?

 

[SpiralFlex]

I got very close to the answer given. But my units were off. What does the question want you to take the constant and mass of the Earth as?

My answer is correct.



Solve.





metres.

 

[Kimyia]

Edit edit: Spiral's got it

 

[SpiralFlex]

Physics dot point book?
This gives you the answer but I don't know if that means the answer is right:
F = GMm/d^2
F is weight force, of which you want 0.01 so its F = 0.01 therefore:
0.01 = 6.67x10^-11 x whatever weight of earth is/d^2
Rearrange to get:
0.01d^2 = 6.67x10^-11 x whatever weight of earth is
Divide RHS by 0.01 and then square root to get: 1.93x10^8.
Let me know if you understand that or not

Edit: wait, that gives 1.99x10^8. Don't know how they got 1.93

For this question you MUST take the mass and radius of the Earth precisely.

 

[SpiralFlex]

Also you didn't minus the radius of the Earth.

 

[Kimyia]

haha, that'd be why. gotcha, thanks Spi

 

[andrew3]

Yep I get it now, thanks for the help.
EDIT: And the question didn't say what value to take the mass of the earth or radius to be.

 

[deswa1]

Sorry, can one of you guys explain where this is wrong:
1. Gravitational acceleration on Earth is 9.81m/s. Therefore if weight is 1/100, gravitational accleration is 0.0981m/s.
2. We know that acceleration=(GmE)/(rE)^2
3. Sub in 0.0981=(6.67x10^-11)x(5.97x10^24)/(r)^2
4. Solve for r and subtract the radius of the Earth and you get 5.73x10^7

Thanks heaps

 

[SpiralFlex]

We're dealing with the force of attraction between two bodies. But, your g assumes that there is a constant gravitational acceleration as you move away from the Earth. This is not the case obviously as with distance, the acceleration of course decreases with accordance with your formula you have provided.

Your formula is also not applicable unless the object is on the surface of the Earth/planet.

Hope that makes sense/helps.

 

[SpiralFlex]

Yep I get it now, thanks for the help.
EDIT: And the question didn't say what value to take the mass of the earth or radius to be.

Safest to take the mass of the Earth as on the BOS standard formula sheet.

Universal gravitational constant, G




Mass of Earth




They don't specifically mention the radius of the Earth so for now it's best to take it as




These values you SHOULD remember.

 

[SpiralFlex]

Also, you using dot point andrew...?

 

[Kimyia]

I think dot point likes to use 5.974x10^24kg for the mass of the Earth and 6378km for the radius.

 

[clementc]

I got very close to the answer given. But my units were off. What does the question want you to take the constant and mass of the Earth as?

My answer is correct.



Solve.





metres.

Can someone explain how this can be right though, even though it matches the answer given? The first line is dimensionally incorrect, as you are equating a force with a mass. The question says that it wants the weight there to be 0.01 of the weight on the earth's surface.

What I did was this:



But yeah anyway that's just what I got =D
deswa1's method also seems okay because he's just finding the gravitational acceleration the question requires - he's not really assuming it's constant, which you can see through the way he uses

.