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zhuklara [117]
3 years ago
7

Find the density of a planet with a radius of 8000 m if the gravitational acceleration for the planet, gp, has the same magnitud

e as the gravitational constant, G (keep the right units for both), where G = 6.67 x 10-11 m3/(kg s2) Hint: Use the expression for the gravitational force and Newton's second law.
Physics
1 answer:
Naya [18.7K]3 years ago
4 0

Answer:

Density = 3 x 10⁻⁵ kg/m³

Explanation:

First, we will find the volume of the planet:

V = \frac{4}{3}\pi r^3\ (radius\ of\ sphere)\\\\V =   \frac{4}{3}\pi (8000\ m)^3\\\\V = 2.14\ x\ 10^{12}\ m^3

Now, we will use the expression for gravitational force to find the mass of the planet:

g = \frac{Gm}{r^2}\\\\m = \frac{gr^2}{G}

where,

m = mass = ?

g = acceleration due to gravity = 6.67 x 10⁻¹¹ m/s²

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

r = radius = 8000 m

Therefore,

m = \frac{(6.67\ x\ 10^{-11}\ m/s^2)(8000\ m)^2}{6.67\ x\ 10^{-11}\ Nm^/kg^2}\\\\m = 6.4\ x\ 10^7\ kg

Therefore, the density will be:

Density = \frac{m}{V} = \frac{6.4\ x\ 10^7\ kg}{2.14\ x\ 10^{12}\ m^3}

<u>Density = 3 x 10⁻⁵ kg/m³</u>

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