# Homework Help: Rate of change of the gravitational force.

1. May 26, 2014

### juanandresrua

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

The space shuttle Endeavor is moving at a speed of 27870 Km / h. At takeoff, its mass is 2, 28 * 10 ^ 6 kg, however, its mass decreases as it uses its fuel: the engine burns liquid oxygen which has a density of 1141 kg / m^3, at a rate of 1340 liters per second. Calculate the rate of change of the gravitational force exerted by earth on the shuttle, at the time the rocket is 20 km in height.

2. Relevant equations

3. The attempt at a solution

I know I have to derivate the formula but I don't get any relation where I can get the dm/dt that results after derivating.

Thanks for help.
So sorry for my English.

Last edited: May 27, 2014
2. May 27, 2014

### Joffan

This is definitely a question where the more you know, the less willing you are to simply calculate an answer. This is made worse by the use of something very close to a real spacecraft name that really worked in a different way that would not allow an answer to be calculated from the information given.

However I think the question wishes to assume the following rather unrealistic scenario:
The craft has steadily accelerated (or, more complicated, has experienced constant thrust) from take-off (speed zero) to reach its current speed over a distance of 20km straight up, losing mass at a constant rate to be calculated from the density of liquid oxygen.

Essentially both m and d are changing with time, so getting $\frac{dF}{dt }$ will require considering both of these.

First thing to do is work out how many seconds have elapsed since take-off. Unless I'm making this more complicated than it needs to be.

Last edited: May 27, 2014
3. May 27, 2014

### Joffan

Let me start a little further back.

Looking at $F=G\left(\frac{mM}{d^2}\right)$, we see that $G$ and $M$ are constant and $d$ and $m$ are changing. So for our purposes, $$F(t)=GM\left(\frac{m(t)}{d(t)^2}\right)$$
We need
$$\frac{dF}{dt}=GM\frac{d}{dt}\left(\frac{m}{d^2}\right)$$

The problem effectively gives us numerical values for $\frac{dm}{dt}$ and $\frac{dd}{dt}$ at the time of interest.

4. May 28, 2014

### juanandresrua

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-xfp1/t1.0-9/10411975_289303221245796_714101699755499906_n.jpg

and like

https://fbcdn-photos-e-a.akamaihd.net/hphotos-ak-xaf1/t1.0-0/10325761_289303194579132_2412401599182272082_t.jpg

Where "Rho" (ρ) is the density of liquid oxygen and "v" is the volume.

Then

https://fbcdn-photos-g-a.akamaihd.net/hphotos-ak-xfa1/t1.0-0/1907645_289303214579130_8064069686441702048_s.jpg

So

Furthermore

https://fbcdn-photos-c-a.akamaihd.net/hphotos-ak-xpa1/t1.0-0/10306544_289303197912465_4206053763258306607_a.jpg

And then I replaced all the data and converted units of dF/dt to N/h (Newton/hour) but I do not get the answer that my teacher gave: -8.9*10^11 N/h

What am I doing wrong?

5. May 29, 2014

### Joffan

- What is $\frac{d}{dt}\left(\frac{1}{d^2}\right)$?

- Recheck $\frac{dm}{dt}$

- What value did you use for $m$? (And indeed $d$, $G$ and $M$?)

- Can you double-check the values given in your opening post, because the accelerations involved here are ridiculous, even for a calculation exercise?