Answer :
The magnitude of the rock's acceleration half-way between two planets is 1.086 m/s².
The given parameters;
- mass of planet, A = 3M
- radius of the planet A, = R
- mass of planet B, = 4M
- radius of planet B, = 2R
- distance between the two planets, r = 8R
The gravitational force half-way (4R) between the two planets is calculated from Newton's universal gravitational law as follows;
[tex]F_g = \frac{Gm_Am_B}{r^2} \\\\F_g = \frac{G(3M\times 4M)}{(4R)^2} \\\\F_g = \frac{12GM^2}{16R^2} \\\\F_g = \frac{12 \times 6.67\times 10^{-11} \times (7.3\times 10^{23})^2}{16 (5.8\times 10^6)^2} \\\\F_g = 7.925 \times 10^{23} \ N[/tex]
The magnitude of the rock's acceleration is calculated from Newton's second law of motion as follows;
F = ma
[tex]a = \frac{F}{m} \\\\a = \frac{7.925 \times 10^{23} }{7.3\times 10^{23}} \\\\a = 1.086 \ m/s^2[/tex]
Thus, the magnitude of the rock's acceleration is 1.086 m/s².
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The acceleration of rock is [tex]1.085 \;\rm m/s^{2}[/tex].
Given data:
The mass of planet A is, 3M.
The radius of planet A is, R.
The mass of planet B is, 4M.
The radius of planet B is, 2R.
The value of M is, [tex]M=7.3 \times 10^{23} \;\rm kg[/tex].
The value of R is, [tex]R = 5.8\times 10^{6} \:\rm m[/tex].
The gravitational force at halfway (4R) between the two planets is,
[tex]F=\dfrac{G(3M)(4M)}{(4R)^{2}}\\F=\dfrac{G \times 12M^{2}}{16R^{2}}[/tex]
Here, G is the universal gravitational constant. Solving as,
[tex]F=\dfrac{6.67 \times 10^{-11} \times 12 \times (7.3 \times 10^{23})^{2}}{16(5.8 \times10^{6})}\\F = 7.925 \times 10^{23} \;\rm N[/tex]
Now, the acceleration of rock is calculated as,
[tex]F = Ma\\7.925 \times 10^{23} = 7.3 \times 10^{23} \times a\\a = 1.085 \;\rm m/s^{2}[/tex]
Thus, the acceleration of rock is [tex]1.085 \;\rm m/s^{2}[/tex].
Learn more about the gravitational force here:
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