Answer:
1.5 hr
16.7
Explanation:
Zero apparent weight means there's no normal force.
Sum the forces in the centripetal direction.
∑F = ma
mg = mv²/r
v = √(gr)
v = √(7.4×10⁶ m × 10 m/s²)
v = 8602 m/s
The circumference of the equator is:
C = 2πr
C = 2π (7.4×10⁶ m)
C = 4.65×10⁷ m
So the period is:
T = C / v
T = (4.65×10⁷ m) / (8602 m/s)
T = 5405 s
T = 1.5 hr
The initial speed is:
v = C / T
v = (4.65×10⁷ m) / (25 h × 3600 s/h)
v = 517 m/s
The speed increases by a factor of:
8602 m/s / 517 m/s
16.7
The new period of the planet at the given radius is 1.5 h.
The speed of the planet increased by a factor of 16.7.
The given parameters;
The speed of the planet at the given radius is calculated as follows;
[tex]mg = \frac{mv^2}{r} \\\\gr = v^2\\\\v= \sqrt{rg} \\\\v = \sqrt{7.4 \times 10^6 \times 10} \\\\v = 8602.3 \ m/s[/tex]
The new period of the planet is calculated as follows;
[tex]T = \frac{2\pi R}{v} \\\\T = \frac{2 \pi \times 7.4 \times 10^6}{8602.3} \\\\T =5,405.72 s \\\\T = \frac{5405.72}{3600} = 1.5 \ hr[/tex]
The initial speed of the planet is calculated as follows;
[tex]v_i = \frac{2\pi R}{T} \\\\v_i = \frac{2\pi \times 7.4\times 10^6}{25 \times 3600} \\\\v_i =516.68 \ m/s[/tex]
The increase in the speed of the planet is calculated as;
[tex]= \frac{v}{v_i} \\\\= \frac{8602.3}{516.68} \\\\= 16.7[/tex]
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