Question

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Answer 1

Answer:

1) The student's weight on Earth is approximately 687.398 N

2) The student's weight on Mercury is approximately 257.85 N

3) The student's weight on the Sun is approximately 19,164.428 N

Explanation:

The mass of the student, m = 70 kg

1) The mass of the Earth, M = 5.972 × 10²⁴ kg

The radius of the Earth, R = 6,371 km = 6.371 × 10⁶ m

The universal gravitational constant, G = 6.67430 × 10⁻¹¹ N·m²/kg²

Mathematically, the universal gravitational law is given as follows;

$F_g =G \times \dfrac{M \cdot m}{R^{2}}$

Therefore, we have;

$F_g=6.67430 \times 10^{-11} \times \dfrac{5.972 \times 10^{24} \cdot 70}{(6.371 \times 10^6)^{2}} \approx 687.398$

$F_g$ = W ≈ 687.398 N

The student's weight on Earth, W ≈ 687.398 N

2) On Mercury, we have;

The mass of Mercury, M₂ = 3.285 × 10²³ kg

The radius of Mercury, R₂ = 2,439.7 km = 2.4397 × 10⁶ m

The universal gravitational constant, G = 6.67430 × 10⁻¹¹ N·m²/kg²

The universal gravitational law is $F_g =G \times \dfrac{M_2 \cdot m}{R_2^{2}}$

Therefore, we have;

$F_g=6.67430 \times 10^{-11} \times \dfrac{3.285 \times 10^{23} \cdot 70}{(2.4397 \times 10^6)^{2}} \approx 257.85$

$F_g$ = W₂ ≈ 257.85 N

The student's weight on Mercury, W₂ ≈ 257.85 N

3) On the Sun, we have;

The mass of the Sun, M₃ ≈ 1.989 × 10³⁰ kg

The radius of the Sun, R₃ ≈ 696,340 km = 6.9634 × 10⁸ m

The universal gravitational constant, G = 6.67430 × 10⁻¹¹ N·m²/kg²

The universal gravitational law is $F_g =G \times \dfrac{M_3 \cdot m}{R_3^{2}}$

Therefore, we have;

$F_g=6.67430 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \cdot 70}{(6.9634 \times 10^8)^{2}} \approx 19,164.428$

$F_g$ = W₃ ≈ 19,164.428 N

The student's weight on the Sun, W₃ ≈ 19,164.428 N