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Likurg_2 [28]
3 years ago
9

Find the magnitude of the gravitational force a 63.5 kg person would experience while standing on the surface of Earth with a ma

ss of 5.98 × 1024 kg and a radius of 6.37 × 106 m. The universal gravitational constant is 6.673 × 10−11 N · m2 /kg2 . Answer in units of N
Physics
1 answer:
Anastaziya [24]3 years ago
3 0

Answer: 3976N

Explanation:

Using the formula for calculating gravitational force between two masses, we have

F = GMm/r^2

Where G is the gravitational constant

M and m are the masses

r is the distance between the masses

F= 6.673 × 10-¹¹ × 5.98 × 10²⁴ × 63.5/ (6.37 × 10^6)^2

F= 2.533×10^16/6.37×10^12

F= 0.3976×10⁴N

F= 3976N

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b. Since the relation \tau=I\alpha hols, being I the moment of inertia, the angular acceleration can be calculated by \alpha=\frac{\tau}{I}. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is I=\frac{1}{2}Mr^2, being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by I_p=m_pr_p^{2}, being m_p the mass of the person and r_p^{2} the distance from the person to the center. Given all of this, we have

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c. Similar equation to b, but changing r_p=2m, so

alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}.

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

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