The mysterious Planet X has a radius of 5.50 x 10^6 m and an acceleration due to gravity of...

Question:

The mysterious Planet {eq}X {/eq} has a radius of {eq}5.50 \times 10^6 {/eq} m and an acceleration due to gravity of 12.0 m/s{eq}^2 {/eq} at its surface. Calculate its mass and average density.

Newton's law of gravitation:

According to this law, the force of gravitational attraction between two masses is directly proportional to the product of the masses and inversely proportional to the square of distance between their centers. Out of the four basic forces in nature, that is, strong nuclear force, weak nuclear force, electromagnetic force and gravitational force , gravitational force is the weakest force.

Answer and Explanation:

A mysterious planet with a radius {eq}R = 5.5 \times 10^6 \ m {/eq} and acceleration due to gravity {eq}g = 12.0 \ m/s {/eq} is given to us.

a. Calculation of mass of this planet

The given value of g applies very close to the surface of the planet.

From Newton's second law, force (F) acting on a body equals the mass of the body (m) times its acceleration (a) in the direction of force.

{eq}F = ma {/eq}

Since {eq}a = g , {/eq}

{eq}F = mg \dots(1) {/eq}

This force F is the gravitational attraction between a small body with mass m and the planet X with mass M.

So, from Newton's gravitation law;

{eq}\begin{align*} F &= \frac{GMm}{(R+ h)^2} \end{align*} {/eq}

where

{eq}\begin{align*} h & \text { is the distance of the mass m from the surface of the planet X.} \\ G & \text { is the universal gravitational constant with } \ G = 6.67 \times 10^{-11} \ Nm^2/Kg^2. \end{align*} {/eq}

Now, {eq}h << R {/eq},

Hence,

{eq}\begin{align*} F &= \frac{GMm}{(R)^2} \dots(2) \end{align*} {/eq}

Equating equation 1 and equation 2 as follows;

{eq}\begin{align*} mg &= \frac{GMm}{(R)^2} \\ M &= \frac { gR^2} { G } \\ M &= \frac { 12 \times (5.50 \times 10^6) ^2 } { 6.67 \times 10^-11 } \\ M &= 5.442 \times 10^{24} \ Kg. \end{align*} {/eq}

b.Calculation of average density :

Density {eq}\rho {/eq} of a planet is mass {eq}M {/eq} of the planet per unit volume {eq}V {/eq} of the planet.

Assuming the planet X to be completely spherical,

{eq}\begin{align*} V &= \frac { 4 \pi R^3 } { 3 } \\ V &= \frac { 4 \pi (5.5 \times 10^6)^3 } { 3 } \\ V &= 6.97 \times 10^{20} \ m^3. \end{align*} {/eq}

Thus, the average density of the planet is calculated as follows;

{eq}\begin{align*} \rho &= \frac { M } { V } \\ \rho &= \frac { 5.442 \times 10^{24} \ Kg } { 6.97 \times 10^20 \ m^3 } \\ \rho &= 7.80 \times 10^3 \ Kg/m^3. \end{align*} {/eq}


Learn more about this topic:

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Newton's Law of Gravitation: Definition & Examples

from Physics: High School

Chapter 7 / Lesson 6
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