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

1. Oct 23, 2016

### Arsen_2872

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
GM/R^2 = g

3. 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!

2. Oct 23, 2016

### TSny

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

3. Oct 23, 2016

### Staff: Mentor

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).

4. Oct 23, 2016

### Arsen_2872

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.

5. Oct 23, 2016

### Staff: Mentor

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

6. Oct 23, 2016

### 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?

7. Oct 23, 2016

### TSny

Yes.

8. Oct 23, 2016

### 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)?

9. Oct 23, 2016

### TSny

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

10. Oct 23, 2016

### Staff: Mentor

11. Oct 23, 2016

### Arsen_2872

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

12. Oct 23, 2016

### Arsen_2872

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

13. Oct 23, 2016

### Staff: Mentor

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

14. Oct 23, 2016

### Arsen_2872

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

15. Oct 23, 2016

### David Lewis

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.

16. Oct 23, 2016

### Staff: Mentor

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.

17. Oct 23, 2016

### David Lewis

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.

18. Oct 23, 2016

### Staff: Mentor

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