Answer:
The acceleration due to gravity near the surface of Ceres is approximately 0.284 m/s²
Explanation:
The given parameters are;
The mass of Ceres = 9.4 × 10²⁰ kg
The radius of Ceres, r = 4.7 × 10⁶ m
By Newton's law of gravitation, we have;
[tex]Gravitational \ force\ on \ the \ spacecraft \ near \ Ceres, \ F =G\dfrac{m_{1}m_{2}}{r^{2}}[/tex]
Therefore;
[tex]Acceleration \ due \ to \ gravity \ g =G\dfrac{m_{1}}{r^{2}}[/tex]
Where;
m₁ = The mass of Ceres = 9.4 × 10²⁰ kg
m₂ = The mass of the spacecraft
G = 6.6743 × 10⁻¹¹ N·m²/kg²
Substituting the values, we get;
[tex]g \ near \ the \ surface \ of \ Ceres = \dfrac{ 6.6743 \times 10^{-11} \times 9.4 \times 10 ^{20}}{(4.7 \times 10^5)^{2}} \approx 0.284 \ m/s^2[/tex]
The acceleration due to gravity near the surface of Ceres ≈ 0.284 m/s²
