How Is the Net Gravity Force Calculated Between Earth, Moon, and a Rocket?

[The Subject]

Homework Statement

The Earth has a mass of 5.98*10²⁴ Kg and the moon has a mass of 7.35*10²² kg. The distance from the centre of the moon to the centre of the Earth is 3.84*10⁸ m. A rocket with a total mass of 1200 kg is 3.0*10⁸ m from the centre of the Earth and directly in between the Earth and the moon. Find the net gravity.

Homework Equations

F_net = $\frac{Gm1m2}{r^2}$

The Attempt at a Solution

The distance from the rocket to the moon is 3.84*10⁸-3*10⁸ = 8.4*10⁷
and the rocket to Earth is 3.84*10⁸ m

E<--------R---->M
let right be positive


F_net=F_{rocket to moon} - F_{rocket to earth}

F_net=$\frac{G(1200)(7.35*10^2)}{8.4*10^7}$-$\frac{G(1200)(5.98*10^2)}{3*10^8}$

F_net=0.83 - 5.3 N

The answer is a negative, so the net force is pulling the rocket towards earth. Knowing this is wrong but what have i done incorrectly?

 

[CWatters]

Perhaps check how far the Lagrange point L1 is from earth?

 

[D H]

The answer is a negative, so the net force is pulling the rocket towards earth. Knowing this is wrong but what have i done incorrectly?

You dropped your units and as a result didn't realize you had a dimensionally incorrect result. Don't treat those dimensioned quantities as just numbers. Here's your error:

F_net=$\frac{G(1200)(7.35*10^2)}{8.4*10^7}$-$\frac{G(1200)(5.98*10^2)}{3*10^8}$

G has dimensions length³·time^-2·mass^-1, or units of m³·s^-2·kg^-1. That 1200, 7.35*10²², and 5.98*10²⁴: Those are masses, in kilograms. The denominators 8.4*10⁷ and 3*10⁸ are lengths, in meters. The units of your expression are thus kg·m²·s^-2 rather than kg·m·s^-2, the units of force.

Homework Equations

F_net = $\frac{Gm1m2}{r^2}$

You didn't use this equation. You didn't square the distance in the denominator.It's a bit more cumbersome to do so, but always carry the units along in a computation. That extra little bit of effort will save your rear end many times over.

 

[haruspex]

and the rocket to Earth is 3.84*10⁸ m

No it isn't. Read the problem statement again.

F_net=$\frac{G(1200)(7.35*10^2)}{8.4*10^7}$-$\frac{G(1200)(5.98*10^2)}{3*10^8}$

As well as the r²→r error D H pointed out, the powers of 10 in the numerator seem to have become truncated.

 

[The Subject]

Fixed up the small errors that you guys mentioned

The Attempt at a Solution

The distance from the rocket to the moon is 3.84*10⁸-3*10⁸ = 8.4*10⁷
and the rocket to Earth is 3.0*10⁸ m

E<--------R---->M
let right be positive


F_net=F_{rocket to moon} - F_{rocket to earth}

F_net=$\frac{(6.67*10^2 Nm^2/kg^2)(1200 kg)(7.35*10^{22} kg)}{(8.4*10^7 m)^2}$-$\frac{(6.67*10^2 Nm^2/kg^2)(1200 kg)(5.98*10^{24} kg)}{(3*10^8 m)^2}$

F_net=0.83 N - 5.3 N

F_net= -4.47 N

The values for the answers still the same cause I didn't write down the equation correctly here in PF (I'm sorry should have proofread more rigorously before posting). I'm still left with a negative value.

Since I made right positive, the F_{rocket to moon} is positive since it is the force of the rocket towards the moon. The F_{rocket to earth} is the force of the rocket in the direction to the earth, so its a negative value. Maybe my reasoning is wrong here?​

 

[D H]

You have close to the right number, and it is pointing in the right direction (toward the Earth is correct). Round that to the precision indicated by the values in the problem and you will have the right answer.

In general it's a good idea to maintain a few more digits of precision in your intermediate calculations than the numbers indicate. You round to the indicated precision in your final answer.

 

[haruspex]

F_net=0.83 N - 5.3 N

F_net= -4.47 N

​

To add to D H's remarks, the sequence above is not valid. 5.3 is only to one decimal place, so you cannot quote two decimal place in the next line.​

 

[CWatters]

I'm still left with a negative value

The L1 point is 3.23 * 10^8 meters from Earth and the spacecraft only 3.0 * 10^8 so negative (towards the earth) appears correct.

 

[haruspex]

The L1 point is 3.23 * 10^8 meters from Earth and the spacecraft only 3.0 * 10^8 so negative (towards the earth) appears correct.

I don't see any connection with Lagrange points. What's your thinking here?

 

[D H]

I don't see any connection with Lagrange points. What's your thinking here?

The net gravitational force at the point in question must be Earthward because it's closer to Earth than is the Earth-Moon L1 point and because the net gravitational force points Earthward at L1.

 

[haruspex]

The net gravitational force at the point in question must be Earthward because it's closer to Earth than is the Earth-Moon L1 point and because the net gravitational force points Earthward at L1.

OK, I see. (I thought it more obvious to consider the net zero point, which is about 3.46*10^8, but either will do.)

 

[CWatters]

OK I admit it :-) I'd forgotten the zero point was a bit further from the Earth than L1.

 

[The Subject]

To add to D H's remarks, the sequence above is not valid. 5.3 is only to one decimal place, so you cannot quote two decimal place in the next line.

Okay, since the significant digits are 1 after the decimal point the answer is -4.5 N

. How about my calculation for F_{rocket to moon}, i should limit it to 2 SD.​

 

[jtbell]

In general it's a good idea to maintain a few more digits of precision in your intermediate calculations than the numbers indicate. You round to the indicated precision in your final answer.

Or do the whole calculation entirely in your calculator without writing down (and rounding off) intermediate results and then keying them in again.

 

[The Subject]

In general it's a good idea to maintain a few more digits of precision in your intermediate calculations than the numbers indicate. You round to the indicated precision in your final answer.

Oh okay thanks D H!

 

[The Subject]

Or do the whole calculation entirely in your calculator without writing down (and rounding off) intermediate results and then keying them in again.

Alright! that's a good tip

 

[jtbell]

You can also avoid keying in duplicate numbers (and possibly doing them inconsistently) by recognizing things that you can factor out:

$$F_{net} = (6.67 \times 10^{-11})(1200)
\left( \frac{7.32 \times 10^{22}}{(8.4 \times 10^7)^2}
- \frac{5.98 \times 10^{24}}{(3 \times 10^8)^2} \right)$$