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3.6×105 km\text{3.6} \times 10^5 \text{ km}

Step-by-step solution

5 steps

To solve this problem, we need to find the point between the Earth and the Moon where the gravitational fields created by each body cancel each other out.

  1. Define variables and the condition for a zero resultant gravitational field.

    Let MEM_E be the mass of the Earth and MMM_M be the mass of the Moon.
    We are given ME=80MMM_E = 80 M_M.
    Let DD be the distance between the centers of the Earth and the Moon, D=4.0×105 kmD = 4.0 \times 10^5 \text{ km}.
    Let xx be the distance from the center of the Earth to the point where the resultant gravitational field is zero.
    The distance from the center of the Moon to this point will then be D−xD - x.
    The gravitational field gg due to a mass MM at a distance rr is given by the formula g=GMr2g = \frac{GM}{r^2}, where GG is the gravitational constant.
    For the resultant field to be zero, the magnitudes of the gravitational fields from the Earth (gEg_E) and the Moon (gMg_M) must be equal: gE=gMg_E = g_M.

  2. Set up the equation for equal gravitational fields.

    GMEx2=GMM(D−x)2\frac{GM_E}{x^2} = \frac{GM_M}{(D-x)^2}
  3. Substitute the masses and simplify the equation.

    We can cancel the gravitational constant GG from both sides.

    MEx2=MM(D−x)2\frac{M_E}{x^2} = \frac{M_M}{(D-x)^2}

    Using the relationship ME=80MMM_E = 80 M_M:

    80MMx2=MM(D−x)2\frac{80 M_M}{x^2} = \frac{M_M}{(D-x)^2}

    We can cancel MMM_M from both sides:

    80x2=1(D−x)2\frac{80}{x^2} = \frac{1}{(D-x)^2}
  4. Solve the equation for the distance xx.

    Take the square root of both sides (we consider the positive root since xx is a distance):

    80x2=1(D−x)2\sqrt{\frac{80}{x^2}} = \sqrt{\frac{1}{(D-x)^2}} 80x=1D−x\frac{\sqrt{80}}{x} = \frac{1}{D-x}

    We know that 80=16×5=45\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}.

    45x=1D−x\frac{4\sqrt{5}}{x} = \frac{1}{D-x}

    Cross-multiply to isolate xx:

    45(D−x)=x4\sqrt{5}(D-x) = x 45D−45x=x4\sqrt{5}D - 4\sqrt{5}x = x

    Group the terms with xx:

    45D=x+45x4\sqrt{5}D = x + 4\sqrt{5}x 45D=x(1+45)4\sqrt{5}D = x(1 + 4\sqrt{5}) x=45D1+45x = \frac{4\sqrt{5}D}{1 + 4\sqrt{5}}
  5. Calculate the numerical value.

    Substitute the value of D=4.0×105 kmD = 4.0 \times 10^5 \text{ km}:

    x=451+45×(4.0×105 km)x = \frac{4\sqrt{5}}{1 + 4\sqrt{5}} \times (4.0 \times 10^5 \text{ km})

    Calculate the numerical value of 451+45\frac{4\sqrt{5}}{1 + 4\sqrt{5}}:

    45≈4×2.236067977≈8.9442719084\sqrt{5} \approx 4 \times 2.236067977 \approx 8.944271908 1+45≈1+8.944271908≈9.9442719081 + 4\sqrt{5} \approx 1 + 8.944271908 \approx 9.944271908 8.9442719089.944271908≈0.900040008\frac{8.944271908}{9.944271908} \approx 0.900040008

    Now, multiply by DD:

    x≈0.900040008×4.0×105 kmx \approx 0.900040008 \times 4.0 \times 10^5 \text{ km} x≈3.600160032×105 kmx \approx 3.600160032 \times 10^5 \text{ km}

    Rounding to two significant figures, as per the given D=4.0×105 kmD = 4.0 \times 10^5 \text{ km}:

    x≈3.6×105 kmx \approx 3.6 \times 10^5 \text{ km}

    The distance from the center of the Earth where the resultant gravitational field is zero is:

    3.6×105 km\text{3.6} \times 10^5 \text{ km}
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