What will the traveler weigh on another planet?

Click For Summary
A traveler weighing 695N on Earth will have a different weight on another planet with a radius three times that of Earth and a mass twice that of Earth. The calculations initially attempted involved using Newton's law of universal gravitation but contained errors, including neglecting the traveler's mass and incorrect exponent handling. A more efficient method suggested using ratios to simplify the calculations, emphasizing the importance of squaring the radius when adjusting for the new planet's dimensions. The correct approach involves comparing the weight equations for both planets, ultimately leading to a clearer understanding of the traveler's weight on the new planet. Accurate calculations are crucial to avoid errors in determining the traveler's weight.
pookisantoki
Messages
44
Reaction score
0
A Space traveler weighs 695N on Earth. What will the traveler weigh on another planet whose radius is 3 times that of Earth and whose mass is 2 times that of earth?

So for this problem I got the mass and radius of the earth
Mass of earth= (5.98 *10^24)
Radius of Eath- (6.38 *10^6)
Since it says mass of Earth is twice of Earth I multipled it by two (5.98 * 10^24)*2= 1.196*10^25
and I multiplied the radius of Earth by three (6.38*10^6)*3=19140000

Then I plugged it into the Newton's law of universal gravitation formula
F=G(m1*m2/R^2)
G=6.674*10^-11

F=(6.674*10^-11)((1.196*10^25)/(19140000^2))=2.898*10^-10
but its wrong what did I do wrong. Thank you!
 
Physics news on Phys.org
pookisantoki said:
F=(6.674*10^-11)((1.196*10^25)/(19140000^2))=2.898*10^-10
but its wrong what did I do wrong. Thank you!
Two problems:
(1) You forgot the traveler's mass.
(2) You messed up the exponent.

Rather than do all that number crunching, just use ratios. Much less chance of error that way.
 
Ratios...How would i set that up?

Would i just put (695*2)/3?
 
pookisantoki said:
Ratios...How would i set that up?

Would i just put (695*2)/3?
Almost, but not exactly. (The radius should be squared.)

Try this:
Weight on Earth = GmM/R^2 = 695 N
Weight on planet = Gm(2M)/(3R)^2 = ?

Compare those two expressions.
 
so I plugged the formula in:
((6.67*10^-11)(695)((5.98*10^24)*2))/(((6.38*10^6)^2)*3)=4540.244
I plugged that in as the answer but it's still wrong...Am i missing a step? or is my calculation wrong?
 
pookisantoki said:
so I plugged the formula in:
((6.67*10^-11)(695)((5.98*10^24)*2))/(((6.38*10^6)^2)*3)=4540.244
I plugged that in as the answer but it's still wrong...Am i missing a step? or is my calculation wrong?
Two new problems:
(1) 695 is the traveler's weight, not his mass. Find his mass.
(2) You multiplied the demoninator by 3 after you squared the radius instead of before.

But this is still the hard way. I'd still play around with the suggestion for using ratios that I gave in my last post:

Hint: Weight on planet = Gm(2M)/(3R)^2 = 2/(3^2) * GmM/R^2
(And you already know what GmM/R^2 equals.)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K