a trial MC question i cant do! help! (1 Viewer)

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1. Andy the astronaut has a weight W on a particular planet. The same planet has a moon with half the mass and half the radius of the planet. His weight on the moon is

2) 2W
b) W
c) 1/2W
d)1/4W

which is it??? eek! and what formula did u use?

thanks guys..greatly appreciated

 

[evette13]

hey mate,
the answer is a (2W)
This is because g = G (mass of planet/(radius of planet)^2)
Sub in the values and u get 2W
(sub in mass of moon as m/2 and likewise with radius of moon)

 

[JumboKHS]

The answer is c) 1/2 W

You dont really need to use the formula for this one because they give you simple numbers but you can use it if you want to.

Can Use F = (Gm1m2 )/d^2

Use made up numbers.

For the person give him a mass of 50kg for example and then make up your own planets.

Planet 1:
mass = 1,000,000
radius = 1,000

Planet 2:
mass = 500,000
radius = 500

plug it into the calculator and devide the answer you get for planet 2 by the answer you get for planet one and you get 2.

 

[DraconisV]

The answer is c) 1/2 W
Use F = (Gm1m2 )/d^2

Use made up numbers.

For the person give him a mass of 50kg for example and then make up your own planets.

Planet 1:
mass = 1,000,000
radius = 1,000

Planet 2:
mass = 500,000
radius = 500

plug it into the calculator and devide the answer you get for planet 1 by the answer you get for planet 2 and you end up with 0.5

Doesn't that give you the gravitational force between the two planets, not the actual gravitation pull of a person towards the planet ?? im confused explain again plz differently

oh i get it now. you use the radius as the d, and the weight of the person as one of the masses and the planet as another, is that right

 

[Xenocide]

errr.....JumboKHS is incorrect here and evette 13 is right. Using the formula F = G (mass of planet/(radius of planet)^2) and letting mass = mass/2 and radius = radius /2 u get f(2) = 2 * f(1). In JUMBOKHS' answer "Another planet has half the mass. Therefore with just this stat his weight on the planet would be 1/4W" is just wrong...mass of planet is proportional to weight. Radius is proportional to weight squared and so halving the radius multiplies the weight by 4.

Using JumboKHS' other method is correct, although long and clunky, but unfortunately he makes the mistake of dividing planet 1's force by planet 2's (as opposed to planet 2 by planet 1 since u want the ratio) and hence gets the inverse of the correct answer.

 

[NT-social]

errr.....JumboKHS is incorrect here and evette 13 is right. Using the formula F = G (mass of planet/(radius of planet)^2) and letting mass = mass/2 and radius = radius /2 u get f(2) = 2 * f(1). In JUMBOKHS' answer "Another planet has half the mass. Therefore with just this stat his weight on the planet would be 1/4W" is just wrong...mass of planet is proportional to weight. Radius is proportional to weight squared and so halving the radius multiplies the weight by 4.

Using JumboKHS' other method is correct, although long and clunky, but unfortunately he makes the mistake of dividing planet 1's force by planet 2's (as opposed to planet 2 by planet 1 since u want the ratio) and hence gets the inverse of the correct answer.

i agree, it should be 2W, hey maybe you can get an extra mark..if u put 2W.. but it mught be too late for that

 

[JumboKHS]

Yep thanks for pointing out my error i will fix it up .

So the logic approach to this question would then be.

Half the mass means 1/2W. Then half the radius means 1/2 multiplied by 4 which gives 2W.

 

[DraconisV]

Half the mass means 1/2W. Then half the radius means 1/2 multiplied by 4 which gives 2W.

Ah ok i get it now. half radius with the inverse square law added in yes i get it now. thx guys

 

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thanks guyz

a little confused at first..but now i get it! thankyou very much..