Question

If the gravitational field strength is 24.8m⋅s−2 at the surface of a planet with radius 2.21×104 km, what is the density of the planet, in kg⋅m−3? Do not write units with the answer.

Answer 1

Given info : gravitational field strength is 24.8 m/s² at the surface of a planet with the radius 2.21 × 10⁴ km.

To find : density of the planet.

Solution : electric field strength, E = GM/R²

Volume of planet, V = 4/3πR³

Let d is the density of the planet.

so, mass of planet is, M = 4/3πR³d

Now electric field strength, E = 4GπRd/3

Here, R = 2.21 × 10^7 m , G = 6.67 × 10¯¹¹ Nm²/Kg², E = 24.8 m/s²

so, 24.8 = (4 × 6.67 × 10¯¹¹ × 3.14 × 2.21 × 10^7 × d)/3

⇒24.8 × 3 = 4 × 6.67 × 3.14 × 2.21 × 10¯⁴ × d

⇒(24.8 × 3 × 10⁴)/(4 × 6.67 × 3.14 × 2.21) = d

⇒ d = 0.4 × 10⁴ = 4000 kg/m³

Therefore the density of the planet is 4000 kg/m³

Answer 2

Answer:

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Explanation:

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