I'm having trouble figuring out how to use the equation with G for gravity

[Arsen_2872]

Homework Statement

You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating how much weaker gravity is 250 km above the Earth's surface.

Homework Equations

GM/R^2 = g

The Attempt at a Solution

I know the value of G is 6.67*10^-4 Nm^2/kg and the mass of the Earth is 5.9*10^24kg. But whenever I plug in the radius values for R, I get a number like: 9.667*10^7m/s^2. which can't be correct.

I can't figure out what I'm doing wrong with my calculations. Any help would be appreciated!

 

[TSny]

Welcome to PF!

I know the value of G is 6.67*10^-4 Nm^2/kg

Look up the value of G to check this. Also, check the units.

 

[fresh_42]

With $G=6.67408 \cdot 10^{-11}\cdot \frac{m^3}{kg\cdot s^2} \; , \; M=5.974 \cdot 10^{24}\cdot kg \,$ and $\, R=( \frac{1}{2} \cdot 12735 + 250)^2 \cdot (10^{3})^2 \cdot m^2$
I calculated $9.1 \frac{m}{s^2}$.

However, one must not forget, that the shuttle itself is in free fall missing the earth, so the value above doesn't play any role (inside the shuttle).

 

[Arsen_2872]

Oh I see what I did wrong with the numerical values. How did you

With $G=6.67408 \cdot 10^{-11}\cdot \frac{m^3}{kg\cdot s^2} \; , \; M=5.974 \cdot 10^{24}\cdot kg \,$ and $\, R=( \frac{1}{2} \cdot 12735 + 250)^2 \cdot (10^{3})^2 \cdot m^2$
I calculated $9.1 \frac{m}{s^2}$.

However, one must not forget, that the shuttle itself is in free fall missing the earth, so the value above doesn't play any role (inside the shuttle).

How did you come up with R? I thought R was the distance between the center of the shuttle to the outside edge of the earth.

 

[fresh_42]

Oh I see what I did wrong with the numerical values. How did you

How did you come up with R? I thought R was the distance between the center of the shuttle to the outside edge of the earth.

Nope. It's the distance between the centers of masses, i.e. the center of earth.

 

[Arsen_2872]

I'm still not sure how to calculate the R values.
Would you just take the distance from the center of the Earth to the shuttle?

 

[TSny]

I'm still not sure how to calculate the R values.
Would you just take the distance from the center of the Earth to the shuttle?

Yes.

 

[Arsen_2872]

So the radius of the Earth is 6.371 *10^6 meters. And the problem says that the shuttle is 2.58*10^5 meters away from the Earth's surface.
Meaning that the shuttle's distance from Earth would be (6.371*10^6 m) - (2.58*10^5 m)?

 

[TSny]

So the radius of the Earth is 6.371 *10^6 meters. And the problem says that the shuttle is 2.58*10^5 meters away from the Earth's surface.

OK, but in the original statement of the problem you have 250 km rather than 258 km.

Meaning that the shuttle's distance from Earth would be (6.371*10^6 m) - (2.58*10^5 m)?

No. R is the distance from the center of the Earth to the shuttle. Imagine starting at the center of the Earth and moving outward until you reached the shuttle. How far would you need to go?

 

[fresh_42]

So the radius of the Earth is 6.371 *10^6 meters. And the problem says that the shuttle is 2.58*10^5 meters away from the Earth's surface.
Meaning that the shuttle's distance from Earth would be (6.371*10^6 m) - (2.58*10^5 m)?

Why minus? The shuttle's height adds to the radius.

 

[Arsen_2872]

Why minus? The shuttle's height adds to the radius.

I don't think I'm picturing the sketch correctly and it's throwing me off.
So, you would add the two values?

 

[Arsen_2872]

Oh I see what I was doing wrong. I figured out the sketch. Thank you for your help! I really appreciate it.

 

[fresh_42]

I don't think I'm picturing the sketch correctly and it's throwing me off.
So, you would add the two values?

R = radius Earth (r) + height (h)

 

[Arsen_2872]

R = radius Earth (r) + height (h)

View attachment 107900

I appreciate the sketch and the clarification. You were very helpful

 

[David Lewis]

...the shuttle itself is in free fall missing the earth, so the value above (9.1 m/s²) doesn't play any role (inside the shuttle).

My understanding is the astronaut (and everything inside the shuttle) must be accelerating towards the center of the Earth at a rate of 9.1 m/s². And the astronaut will weigh 91/98 of what he weighs on Earth, if we define weight as the gravitational force acting upon him.

 

[fresh_42]

My understanding is the astronaut (and everything inside the shuttle) must be accelerating towards the center of the Earth at a rate of 9.1 m/s². And the astronaut will weigh 91/98 of what he weighs on Earth, if we define weight as the gravitational force acting upon him.

Yes, but as the scale (the shuttle) is equally accelerated, you cannot measure his weight. The system itself is in free fall plus a tangential velocity component, that keeps it from hitting the earth.
Here's (ff.) is a nice discussion on the subject.

 

[David Lewis]

…as the scale (the shuttle) is equally accelerated, you cannot measure his weight.

The astronaut’s weight (under my definition) will be his mass times acceleration. But if it’s the reading on a scale (a different definition) then he is weightless.

 

[fresh_42]

The astronaut’s weight (under my definition) will be his mass times acceleration. But if it’s the reading on a scale (a different definition) then he is weightless.

Yes, but the OP asked ...

why astronauts feel weightless

... so it's not about mass. Since they are in free fall, the actual value of their acceleration doesn't matter.