1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: What force needed to launch an object from moon to Earth

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data

    An object possesses 61.14 kg mass.

    1. How much energy in kilojoules would be necessary so as to project the object from the surface of the moon to the Earth's atmosphere at the Karman Line of 100km Earthly elevation, at such a speed that the total travel time between lunar surface and Earth atmosphere would be 12 hours?
    2. What would be the speed of the object as it collides with the Earth's atmosphere at the Karman line?

    gravitational pull on the moon: 1.622 m/s^2
    Earth's gravity: 9.80665 m/s^2
    mass of the moon: 7.34767309 × 10^22 kilograms
    Mass of the Earth: 5.972 * 10^24 kg
    radius of the moon at surface: 1,737.287 km
    radius of the Earth at Karman Line: 6,471 km
    total distance between Earth's center and moon center: 392,580.569 km
    distance between moon surface and Earth at Karman line (distance to be traversed in 12 hrs): 384,372.282 km

    2. Relevant equations

    3. The attempt at a solution

    I'm thinking that I'll have to construct a function that incorporates both the moon's and the earth's gravity. This gets restated algebraically so that the function's product is time. Then I integrate from 0 to however many seconds there are in an hour.
  2. jcsd
  3. Dec 30, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Why an hour? The trip is supposed to take 12 hours.

    BTW, don't you know how many seconds there are in 1 hour without calculating?

    Hint: How fast do you need to go to escape lunar gravity?

    Is this speed slower or faster than the average speed of the trip between the moon and the earth, given the required time for the trip?
  4. Dec 30, 2015 #3
    Basically you need to keep in mind that the acceleration and thus the velocity at any distance x (from the center of the moon) is a function of x. This could be derived from GMm/x2. Once you can find the velocity in terms of x you can integrate to find the distance traveled for 12 hrs with an additional parameter 12v0 and equate it to 384,372.282 km to find v0.

    This could be complicated computationally.
  5. Jan 6, 2016 #4
    There are two forces of gravitational attraction: that of the Earth and the moon. At a certain distance between these bodies, their opposing gravitational forces are balanced. It would be easier to calculate the necessary force of lunar launching by dividing the span between the Earth and Moon into two sections therefore: one where the gravity of the moon is stronger, and the other where that of the Earth is stronger. The object must first be launched with sufficient force to have it reach this balancing point, and then further must possess enough residual velocity to have it enter the Earth's gravitational zone such that the object will encounter the Karman Line within the 12 hours time limit.

    It would seem that the first step is to calculate precisely where that point of gravitational balance is. Using the gravity formula of $$Big \text {G}$$
  6. Jan 6, 2016 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's not apparent to me that there is a benefit in breaking it into those two parts. The same equation will describe the motion before and after that point, with no discontinuity.
  7. Jan 6, 2016 #6
    What would this equation look like?
  8. Jan 6, 2016 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Suppose it is at distance x from the moon. What are the forces acting on it? What is the net force? What is the acceleration?
    Edit: probably most convenient to measure from the moon's centre.
  9. Jan 6, 2016 #8
    First I take the total distance that is to be traversed (384,372,282 meters), and divide it by the number of seconds in 12 hours (43,200):$$\frac{384,372,282_m}{43,200_s} = 8,897.50652778_{m/s}$$

    This is the average velocity which the object will have to possess throughout its journey.

    Then I create a function that accounts for position along the flightpath based off of time, yes?

    There are two forces; the gravitational pull of the Moon, and that of the Earth.

    The net force will depend upon its position upon the imaginary flightpath (which I simplify to a straight line). Close to the Moon, the Moon's gravity predominates, whereas at most other locations it's that of Earth.

    If I were to consider the Moon in isolation, its gravitational pull is a function of position along the line from 1,736,482 meters (radius of the Moon, or starting point), to 386,108,764 meters (location of Karman Line, determined by adding lunar radius by the distance already mentioned as regards between the Moon surface and Karman Line). Its force would consist of a function that looked like this:$$F = G \frac{Mm}{r^2} $$ $$F_{Moon} = -\left(G \right) \left( \frac{(7.34767309 * 10^{22}{}_{\text{kg}})(61.14_{\text{kg}})}{r^2} \right)$$

    Or:$$F_{Moon} = -\left( \frac{2.9982419*10^{14}}{r^2} \right)J$$
    I make the sign of the Moon's force negative because it is pulling towards the opposite direction of the flightpath

    Next I plug in the values for Earth's force, which are positive, because they attract towards the direction of the flightpath, and in a similar fashion simplify to yield:$$F_{Earth} = \left( \frac{2.436894*10^{16}}{r^2} \right)J$$

    Thus the sum of forces is for my object at any given point is:$$F _{\text{for any r}} = \left( \frac{2.436894*10^{16}}{r^2} \right)-\left( \frac{2.9982419*10^{14}}{r^2} \right)J$$

    Or to combine the fractions algebraically and simplify the numerator, I thus get:$$F _{\text{for any r}} = \left( \frac{2.4069116*10^{16}}{r^2} \right)J$$

    Now...this cannot be right! If someone were standing on the moon, they would just be sucked up into the Earth. So I guess I would have to redevise my function of Force with position as an input, by having the relative forces of the Earth and Moon apply not equally across the entire flightpath, but rather would have different values for different positions along the path.

    How would that be written?
  10. Jan 6, 2016 #9


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Unlikely that this will be useful.
    I strongly encourage you to resist plugging in numbers until the latest possible point. Many advantages in staying symbolic.
    Where J represents...?
    Using the same r for both distances is not going to lead to a happy ending.
  11. Jan 6, 2016 #10
    To account for the simultaneously increasing proximity and distance of the Earth and Moon respectively along the traversing, and making the Moon's center be the frame of reference;
    1. ##r_{\text{Moon}}## I set to be the distance along the flightpath from the moon's center, so at the initial condition it is located at the Moon's radius away from that center.
    2. I added together the distance that the flightpath will take with the Earth's radius:$$384,372,282_m + 6,471,000_m = 390,843,282_m$$
    3. Setting this number as what the distance is from the Earth's center at the beginning, I then subtract from this number the ##r_{\text{Moon}}## value, save the Moon's radius (1,737,287 m), and thus determine the Earth's distance using the reference of the moon:$$r_{\text{Earth}} = 390,843,282_m - (r_{\text{Moon}} - 1,737,287_m)$$

    These different variables are entered into different fractions to represent the different forces from each of the bodies thusly;$$F_{\text{Moon}}=-\left(\frac{2.9982419∗10^{14}}{r_\text{Moon}{}^2}\right)\text{Joules}$$

    These should work as functions to determine force at any given point located away from the lunar frame of reference, yes?
  12. Jan 6, 2016 #11


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Since when has force, gravitational or otherwise, been measured in joules?
  13. Jan 7, 2016 #12


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, but please heed my request not to plug in numbers yet. Rewrite it symbolically.
    What is the relationship between rEarth and rMoon?
  14. Jan 7, 2016 #13


    User Avatar

    Staff: Mentor

    A question if I may? What course does this question come from?

    The reason I ask is that the time criterion (you want the traversal from Moon to Karman Line to take a specific amount of time) turns the problem into one where you need to determine the time of flight of an object on a trajectory which is for all intents and purposes a body on an orbit in a three-body system.

    If the course involves applying numerical methods to solve physical problems then your approach can be straightforward. Otherwise I fear that the math will prove to be rather nasty indeed.
  15. Jan 9, 2016 #14
    Let me begin by sincerely thanking all of you for your comments and direction. It is proving most valuable in raising my level of competence as regards the physics and math involved.

    Steamking: Thanks...I now surmise (I believe correctly) that force is measured in Newtons, whereas energy is measured in Joules. I got Joules because I didn't square the meter unit in the denominators, because I thought that you can't do such things as compute or exponentially increase or diminish the value of units, whereas this can be done with scalars, but I must have been wrong about this. I reason that energy is force applied to a given distance, or is this work? What is the arcane distinction then between energy and work? Is work the kinetic species of energy, whereas energy measured in Joules could have many different varieties? Finally: if I do arrive at some number of Newtons, how do I convert this to Joules?

    haruspex: You give sound advice regarding the utility of remaining symbolic initially. So much so that I realize I need to drop the lunar radius component latterly subtracted in parantheses from the ##r_{\text{moon}}## term. I also realized that the calculation should involve simply a distance between the centers of mass already mentioned. So to answer your question in symbolic terms, the relationship of ##r_{\text {Earth}}## to ##r_{\text {Moon}}## is:
    ##r_{\text {Earth}}~ =## (distance in meters from Earth COM to Moon COM) - ##r_{\text {Moon}}##.

    gneill: the whole system is not meant to be treated very realistically, but rather is grossly simplified such that the moon and Earth are not rotating, are stationary in respect to each other, and that there are no other systems astrophysically. Also, the path of travel is in a straight line. Perhaps I neglected to mention these simplifying assumptions previously. You bring to my attention an interesting detail though. Would however the force necessary in my very simplified system be less than the necessary force of a real space vehicle in a real system, or the same (when all outside gravitational forces are disregarded)?

    Now I come to resume my struggle to solve this problem...

    At any given point along the path, the total gravitational force exerted upon this object will be considered as:$$F_{\text {Moon}} - F_{\text {Earth}} = F_{\text{gravity}}$$
    ...using the lunar center of mass as the frame of reference. Given the much greater mass of the Earth compared to the moon, over this path there will be a far greater net force exerted by the Earth. And so, if I was to simply integrate across the entire distance, I would end up with a negative number, indicating that no force would be necessary to fall to the Earth from the Moon's surface, which is obviously wrong. I only have to make the object overcome the gravitational pull of the Moon, and then can fall to the Earth after reaching a point of equilibrium betwixt these differing gravitational forces. To calculate the force necessary to overcome the lunar gravity and reach this point of equilibrium, I must first compute where this point is on the flightpath. To do so, I set the gravitational force equal to 0 in the former equation:
    $$F_{\text {Moon}} - F_{\text {Earth}} = 0$$
    And then add ##F_{\text {Earth}}## to both sides, thus generating:
    $$F_{\text {Moon}} = F_{\text {Earth}}$$
    Next I restate this equation by putting in all the variables of these force functions:
    $$G \left(\frac {(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}^2} \right)= G \left(\frac {(mass_{\text{earth}})(mass_{\text{object}})}{r_{\text{earth}}^2} \right)$$
    This equation may however be algebraically simplified by eliminating the numbers that occur on both sides of the equation to render:
    $$\frac {mass_{\text{moon}}}{r_{\text{moon}}^2} = \frac {mass_{\text{earth}}}{r_{\text{earth}}^2}$$
    restating ##r_{\text{earth}}## in terms of ##r_{\text{moon}}##yields (with the phrase: "COMCOM" indicating the distance between the centers of mass for the Earth and Moon):
    $$\frac {mass_{\text{moon}}}{r_{\text{moon}}^2} = \frac {mass_{\text{earth}}}{((\text{COMCOM}) - r_{\text{moon}})^2}$$
    A little algebra and simplification is applied to isolate ##r_{\text{moon}}## thusly:
    $$\frac {\text{COMCOM}}{\sqrt{\frac{\text{mass}_{\text{Earth}}}{\text{mass}_{\text{Moon}}}} +1} = r_{\text{balance point}}$$
    Next, known and aforementioned values are plugged into the formula to thus give us (these values were listed above in the initial entry of this thread):
    $$\frac {\text{392,580,569_m}}{\sqrt{\frac{5.972 * 10^{24}_{}\text{kg}}{7.34767309 × 10^{22}_{}\text{kg}}} +1} = r_{\text{balance point}}$$
    Which computes to: ##39,197,693.5716_m## away from the Moon's COM. This will form the upper boundary of my integration, with the lower end being the starting point at the lunar surface or ##1,737,287_m##.

    Taking my previously established formula of gravitational force at any point ##F_{\text {Moon}} - F_{\text {Earth}} = F_{\text{gravity}}##, I incorporate all the relevant variables to render it into a complete form viz:
    $$\left(G \left(\frac {(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}^2} \right)\right) - \left( G \left(\frac {(mass_{\text{earth}})(mass_{\text{object}})}{(\text{COMCOM} - r_{\text{moon}})^2} \right)\right) = F_{\text{gravity}}$$

    Now to integrate...
    I'll integrate first for the moon's gravity, and then the Earth's. Represented symbolically, it appears thusly:
    $$\left(\int_{\text{start}}^{\text{balance point}}\!G \left(\frac {(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}^2} \right)dr\right) - \left(\int_{\text{start}}^{\text{balance point}}\! G \left(\frac {(mass_{\text{earth}})(mass_{\text{object}})}{(\text{COMCOM} - r_{\text{moon}})^2} \right)dr\right) = F_{\text{to reach balance point}}$$
    The constants in the numerators and G can all be multiplied together and taken outside the integral to be multiplied by the result later:
    $$\left((G)(mass_{\text{moon}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac {1}{r_{\text{moon}}^2} \right)dr\right) - \left((G)(mass_{\text{earth}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac {1}{(\text{COMCOM} - r_{\text{moon}})^2} \right)dr\right) = F_{\text{to reach balance point}}$$
    The integrals themselves are fairly simple - being quickly looked up in a table to be;
    $$(G)(mass_{\text{moon}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac {1}{r_{\text{moon}}^2} \right)dr = -\frac{(G)(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}}_{\text{balance point}} +\frac{(G)(mass_{\text{moon}})(mass_{\text{object}})}{r_{\text{moon}}}_{\text{start}}$$
    for the first, and guessing and checking to generate:
    $$(G)(mass_{\text{earth}})(mass_{\text{object}})\int_{\text{start}}^{\text{balance point}}\! \left(\frac{1}{(\text{COMCOM} - r_{\text{moon}})^2} \right)dr = \frac{(G)(mass_{\text{earth}})(mass_{\text{object}})}{-r_{\text{moon}} + COMCOM}_{\text{balance point}} - \frac{(G)(mass_{\text{earth}})(mass_{\text{object}})}{-r_{\text{moon}} + COMCOM}_{\text{start}}$$
    for the second.

    Now that I've determined what these formulas of integration symbolically are, I plug in the values for ##F_{\text{moon}}## and compute:
    $$(G)(mass_{\text{moon}})(mass_{\text{object}}) = (6.67408 *10^{-11}_{~~~~~~~m^3 s^{-2}\text{kg}^{-1}})(7.34767309 * 10^{22}_{~~~\text{kg}})(61.14_{\text{kg}}) = 2.9982419 * 10^{14}_{~~~m^3 s^{-2}\text{kg}}\Rightarrow$$
    $$ -\frac{2.9982419 * 10^{14}_{~~~m^3 s^{-2}\text{kg}}}{39,197,693.5716_m}_{\text{balance point}} +\frac{2.9982419 * 10^{14}_{~~~m^3 s^{-2}\text{kg}}}{1,737,287_m}_{\text{start}} = -7,649,026.30726_{\text{Joules}} + 172,581,841.688_{\text{Joules}} = 164,932,815.38_{\text{Joules}} = F_{\text{moon}}$$
    Next I similarly plug in and compute values for ##F_{\text{earth}}##:
    $$(G)(mass_{\text{earth}})(mass_{\text{object}}) = (6.67408 *10^{-11}_{~~~~~~~m^3 s^{-2}\text{kg}^{-1}})( 5.972 * 10^{24}_{~~~\text{kg}})(61.14_{\text{kg}}) = 2.436894 * 10^{16}_{~~~m^3 s^{-2}\text{kg}}\Rightarrow$$
    $$ \frac{2.436894 * 10^{16}_{~~~m^3 s^{-2}\text{kg}}}{-39,197,693.5716_m + 392, 580, 569_m}_{\text{balance point}} -\frac{2.436894 * 10^{16}_{~~~m^3 s^{-2}\text{kg}}}{-1,737,287_m +392, 580, 569_m}_{\text{start}} = 68,959,029.5855_{\text{Joules}} - 62,349,645.2985_{\text{Joules}} = 6,609,384.287_{\text{Joules}} = F_{\text{earth}}$$
    Now that I have integrated both functions of ##F_{\text{moon}}## and ##F_{\text{earth}}## as well as computed their values over the range of the limited integral, I simply need to subtract one value from another according to the aforementioned equation of ##F_{\text {Moon}} - F_{\text {Earth}} = F_{\text{gravity}}##....
    $$164,932,815.38_{\text{Joules}} - 6,609,384.287_{\text{Joules}} = 158,323,431.093_{\text{Joules}}= F_{\text{gravity}}$$

    ...Which is the necessary energy to reach this point of equilibrium.

    Ladies and gentlemen: I'm going to pause here for now, due to the following reasons; I wish to give you all an opportunity to scrutinize my work thus far and determine if there are any necessary corrections and thus forestall compounding errors and/or building upon faulty intellectual/mathematical foundations as it were, but also because I'm tired.:wink: So here is my attempt at a solution for now. Stay tuned for the next episode...
  16. Jan 9, 2016 #15


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I don't know why you would think this.

    For example, the area of a square is equal to the square of the length of the sides.

    If you have a square which measures 1 m on a side, the area is A = (1 m)2 = 1 m2

    Different units can be multiplied together or divided; only like units can be added or subtracted.

    Force is force and energy is energy. There is no conversion between newtons and joules, just like there is no conversion between newtons and meters, for instance.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted