Force of Earth's gravity on a spacecraft

[fattydq]

Calculate the force (in Newtons) of Earth's gravity on a spacecraft 12900 km ABOVE the Earth's surface if its mass is 1410 kg.Give 3 sig figs.Be careful with your units!

Again, here I know force is mass times acceleration, so would it just be 1410 times 9.8? The thing that worries me is this seems way too simple for the type of problems we've been doing, haha, I feel like I'm missing something?

 

[fattydq]

And where does the fact that it's 12900 km above Earth's surface come into play here??

 

[D H]

Again, here I know force is mass times acceleration, so would it just be 1410 times 9.8?

No. That would be wrong.

And where does the fact that it's 12900 km above Earth's surface come into play here??

Hint: Newton's law of gravitation.

 

[fattydq]

No. That would be wrong.


Hint: Newton's law of gravitation.

Hmm I know the basics behind that law but don't see how I can numerically use that principal to solve this?

 

[fattydq]

I honestly thought that law came into effect in the form of an equation when you're working with two different masses and whatnot...how would I apply it in this situation?

 

[fattydq]

Unless...does the mass of the entire Earth come into play here? That wouldn't really make much sense, because that isn't really known is it? Again what are the two unique masses I'm supposed to use here?

 

[D H]

Unless...does the mass of the entire Earth come into play here?

Correct.

That wouldn't really make much sense, because that isn't really known is it?

Where did you ever get this idea?

 

[Monocerotis]

Fg = G(m₁)(m₂)/d²

G = 6.67x10^-11
m₁ = mass of ship
m₂ = mass of earth
d = distance (note: squared)

 

[fattydq]

Then what is the known mass of the earth? Because I just searched it and got 4 different (similar) answers...

 

[fattydq]

5.9742 × 10^24 seems to be the most popular, yes?

 

[fattydq]

I got 3380000000 and my units would be Newtons, is this correct?

 

[D H]

Monocerotis,

1. Please don't give too much help here. Our job is to help students do their own homework, not do it for them.

2. G=6.67x10^-11 what? What units, to be specific.
G is not a unitless constant. Solar system dynamists are just as likely to use G=1.4879×10^-34 AU³/day^²/kg as they are to use G=6.674×10^-11 m³/s²/kg.

 

[D H]

I got 3380000000 and my units would be Newtons, is this correct?

Two problems here.
1. Units.
2. The distance is to the center of the Earth, not the Earth's surface.

 

[fattydq]

Monocerotis,

1. Please don't give too much help here. Our job is to help students do their own homework, not do it for them.

2. G=6.67x10^-11 what? What units, to be specific.
G is not a unitless constant. Solar system dynamists are just as likely to use G=1.4879×10^-34 AU³/day^²/kg as they are to use G=6.674×10^-11 m³/s²/kg.

So my final units wouldn't be Newtons? I know the G constant is in m^3 kg^-1 s^-2 but since it asks for the final answer in Newtons I'm assuming that's what the end product will be?

 

[fattydq]

Two problems here.
1. Units.
2. The distance is to the center of the Earth, not the Earth's surface.

Well how the heck am I supposed to know the distance to the center of the Earth when it would surely vary at different poles/locations on earth?

And how can the units be converted to Newtons if what I figured out isn't in Newtons? Because the original problem says calculate the force in Newtons...

 

[D H]

Well how the heck am I supposed to know the distance to the center of the Earth when it would surely vary at different poles/locations on earth?

Ignoring the 21 km difference between the equatorial and polar radius is a heck of a lot better than ignoring the radius of the Earth altogether.

And how can the units be converted to Newtons if what I figured out isn't in Newtons? Because the original problem says calculate the force in Newtons...

What are the units for G? Is this compatible with the other units you are using?

 

[fattydq]

Ignoring the 21 km difference between the equatorial and polar radius is a heck of a lot better than ignoring the radius of the Earth altogether.What are the units for G? Is this compatible with the other units you are using?

The units for force is Newtons normally, and gravity is a force that's being calculated here, I still don't see what you're getting at. I appreciate the help, I hope I don't sound ungrateful, it just seems you're answering my questions with more complex questions that I can't answer.

 

[D H]

You used G=6.67x10^-11. What are the units associated with that number?

 

[D H]

Suppose you had been told the satellite was 8016 miles above the surface of the Earth instead of 12900 km. How would you have solved the problem given this information?