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Couples of physics questions .

  1. Mar 30, 2004 #1
    Couples of physics questions.....

    i have couples of physics question..can someone help..thanks in advance...

    1.Synchronous communications satellites are placed in a circular orbit that is 1.57 x 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?

    2.At a distance H above the surface of a planet, the true weight of a remote probe is 14.7 percent less than its true weight on the surface. The radius of the planet is R. Find the ratio H/R.

    3.While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.419. The pushing force is directed downward at an angle theta below the horizontal. When theta is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find that value of theta.

    for number1 i know i'm suppose to used F=ma and a=GM/R^2 ......but i dunno what's R...so i don't know what to do...

    #2..i'm just lost..don't know what to do...

    #3..dymanic_Friction_Coeficient * [m*g + F_applied*sin(theta)]=F_applied*cos(theta) ....is this the forumla for it..if it's..i don't know what number goes to which variables...please help...thanks
  2. jcsd
  3. Mar 30, 2004 #2
    1. The gravitational force between two masses is:
    [tex]F_G = G\frac{m_1m_2}{d^2}[/tex]
    In our case, m1 can be the Earth's mass, m2 can be the satellite's mass, and d is the distance between them which is also the Earth's radius plus the height of the satellite above the surface of it. And as you know F = ma, so to get the acceleration divide the force by the satellite's mass. The radius of Earth should already be given by your textbook, it is assumed you know it unless specified otherwise.

    2. The weight of the object on the planet's surface is:
    [tex]F_{G1} = G\frac{Mm}{r^2}[/tex]
    At a distance H above the surface, it is:
    [tex]F_{G2} = G\frac{Mm}{(r + H)^2}[/tex]
    You know what the ratio between the weights is:
    [tex]\frac{F_{G1}}{F_{G2}} = \frac{{(r + H)}^2}{r^2} = \frac{100\%}{85.3\%}[/tex]
    [tex]\frac{r + H}{r} = 1 + \frac{H}{r} = \sqrt{\frac{100\%}{85.3\%}} = 1.083[/tex]
    Last edited: Mar 30, 2004
  4. Mar 30, 2004 #3

    Doc Al

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    Staff: Mentor

    R is the distance from the center of the Earth. In this case, R = radius of Earth (look it up!) plus the height above the surface (given as 1.57 x 107 m).
    The weight is the force of gravity at each point, given by F = GMm/R^2 (here R = distance from center of planet). You need to compare the weights at both points: at the surface (distance = R) and at the height H (distance = R + H). (Hint: use ratios--G, M, & m will cancel.)
    Since velocity is constant, the object is in equilibrium: Forces in the x and y directions add to zero. The applied force has two components: Fx = F cos θ ; Fy = F sin θ (downward).

    The forces on the object in the y direction are: Normal force (up); weight (down); applied force (down). They must balance:

    N = mg + F sin θ

    The forces on the object in the x direction are: Appplied force (to the right, say), friction (to the left). They too must balance:

    F cos θ = N μ

    Combine these equations eliminating "N", then rearrange so that "F" is on one side; see what conclusions you can draw.
    Last edited: Mar 30, 2004
  5. Mar 30, 2004 #4
    so...m2 is 1.57 x 107 m..then what's m1..mass of earth is that supposed to be given in the text too...and number 2..> 1.083..what's that...sorry to ask many questions..but i just don't know...oh my goodness i'm going crazy..with this physics stuff...
    Last edited: Mar 30, 2004
  6. Mar 30, 2004 #5
    No, m2 is not 1.57 x 107 m. That is the height of the satellite above the Earth's surface, so the total distance from the center of Earth is R + H, and R should be given in your textbook. You don't have m2, the mass of the satallite, but you don't need it either in order to find the acceleration of it.

    And [tex]1.083 = \sqrt{\frac{100\%}{85.3\%}}[/tex]
  7. Mar 30, 2004 #6
    it's that the ratio for H/R ???
    Last edited: Mar 30, 2004
  8. Mar 31, 2004 #7

    Doc Al

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    Staff: Mentor

    No. That equals 1 + H/r. See Chen's last equation in post #2. You can solve for H/r, right?
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