 # A space traveler weights 540 N on earth. what will the traveler weigh on another planet whose radius is three times that of earth and whose mass is twice that of earth ? Falak Kinney 2021-01-02 Answered
A space traveler weights 540 N on earth. what will the traveler weigh on another planet whose radius is three times that of earth and whose mass is twice that of earth ?
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Here is what you need to know
W=MG
Weight is equal to mass times (big) G
Normally we call the gravity on earth little g and it is 9.8 aprox 10 the way we calculate this is with the following equation where one the masses is the mass of the earth the other is the mass ofthe weight of the object you want and r is the radius of the earth.
$G\frac{{M}_{1}{M}_{2}}{{r}^{2}}$
And where G is a constant of $6.7×{10}^{-11}$
So now we analyze if the radius of the new planet is tripled then the weight of the person will have to be multiplied by $\frac{1}{{r}^{2}}$ then r is 3 times that of earth so we must multiply his weight by $\frac{1}{9}$
But since the mass of the new planet is twice that of the earth and the mass is in the numerator we must multiply the weight of the person by 2 also
So in order to figure out the answer we multiply his weight on earth times 2/9 which describes the new planet which comes out to
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Step 1

Concept:

As we know the core of classical mechanics is Newton's three laws of motion which the basic classical laws describing motion. As we know the Newton's first law assigns that if the net force on an object s zero, an object originally at rest remains at rest, and an object in motion remains in motion in a straight line with constant velocity.

Also, we know that Newton's second law assigns that the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass

$\sum F=ma$

As we know Newton's third law assigns that whenever one object exerts a force on a second object, the second object always exerts a force on the first object which is equal in magnitude but opposite in direction

${F}_{AB}=-{F}_{BA}$

Step 2

Concept:

As we know the weight refers to the gravitational force on an object, and is equal to the product of the object's mass m and the acceleration of gravity g

${F}_{G}=mg$

The force which is a vector, can be considered as a push or pull. As we know the friction that object exerts on the other can be written as

${F}_{friction}={\mu }_{k}{F}_{N}$

Where, ${F}_{N}$ is the normal force (the force each object exerts on the perpendicular to their constant surfaces).

As we know Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to square to the square of the distance between them:

${F}_{G}=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}$

The direction of this force is along the line joining the two particles, and the force is always attractive. It is this gravitational force that keeps the Moon revolving around the Earth, and the planets revolving around the Sun.

Step 3

Calculation:

Solve for Earth case:

As we mention before in the concept session, Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to square to the square of the distance between them:

${F}_{Earth}=G\frac{{m}_{Earth}{m}_{2}}{{r}_{Earth}^{2}}$

Solve for planet case:

As we mention before in the concept session, Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force proportional to the product of their masses and inversely proportional to square to the square of the distance between them:

${F}_{planet}=G\frac{{m}_{planet}{m}_{2}}{{r}_{planet}^{2}}$

Divide both equations together, then we get

$\frac{{F}_{planet}}{{F}_{Earth}}=\frac{G\frac{{m}_{planet}{m}_{2}}{{r}_{planet}^{2}}}{G\frac{{m}_{Earth}{m}_{2}}{{r}_{Earth}^{2}}}$

$=\frac{\frac{3{m}_{Earth}}{4{r}_{Earth}^{2}}}{\frac{{m}_{Earth}}{{r}_{Earth}^{2}}}$

$=\frac{3}{4}$

So, the weight of the traveler on the planet will be

${F}_{Planet}=\frac{3}{4}{F}_{Earth}$

$=\frac{3}{4}×540N$

=405  N