Calculating Projectile Distance on Pluto

  • Thread starter Thread starter y90x
  • Start date Start date
  • Tags Tags
    Pluto Projectile
Click For Summary

Homework Help Overview

The discussion revolves around calculating the horizontal distance a rock travels when thrown from a height on Pluto, considering its gravitational acceleration. The problem involves projectile motion and the application of gravitational equations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore how to determine gravitational acceleration on Pluto and discuss the relevant formulas involving the gravitational constant. There are attempts to clarify the relationship between mass and gravitational force.

Discussion Status

The discussion is ongoing, with participants questioning the assumptions about mass and gravitational acceleration. Some guidance has been offered regarding the use of gravitational formulas, but there is no clear consensus on the correct value of gravitational acceleration.

Contextual Notes

There is confusion regarding the mass of the rock and the mass of Pluto, as well as the implications of these values for calculating gravitational acceleration. Participants are working with limited information and are trying to clarify these concepts.

y90x
Messages
47
Reaction score
0

Homework Statement



An astronaut throws a rock at 12 m/s at an angle of 25° above horizontal from a height of 3.25 m above the surface. How far horizontally does the rock travel before hitting the surface of Pluto?

Pluto has a mass of 1.27 E22 kg
And a radius of 1.14E6 m

Homework Equations



(Projectile equations )
d=vo(t) + .5a(t^2)
Dx=vx (t)

The Attempt at a Solution


I know how to solve projectiles , except how do I find gravitational acceleration for Pluto ? [/B]
 
Physics news on Phys.org
y90x said:
how do I find gravitational acceleration for Pluto ?
Do you know a formula involving the gravitational constant G?
 
haruspex said:
Do you know a formula involving the gravitational constant G?

Fg = (Gm)/r^2 ?
 
y90x said:
Fg = (Gm)/r^2 ?
Not quite. If you have a force on the left then you need two masses on the right. Or you can have an acceleration on the left and just the attracting mass on the right.
 
haruspex said:
Not quite. If you have a force on the left then you need two masses on the right. Or you can have an acceleration on the left and just the attracting mass on the right.

Fg=(G(m1)(m2))/r2
But then , where would the other mass come from ? Only one mass is given
 
y90x said:
Fg=(G(m1)(m2))/r2
But then , where would the other mass come from ? Only one mass is given
You only need to know the vertical acceleration of the rock. According to the formula above, the force is proportional to its mass, so the acceleration is independent of its mass.
 
haruspex said:
You only need to know the vertical acceleration of the rock. According to the formula above, the force is proportional to its mass, so the acceleration is independent of its mass.

I don’t understand
 
y90x said:
I don’t understand
Let the mass of the rock be m. What is the gravitational force on it?
 
haruspex said:
Let the mass of the rock be m. What is the gravitational force on it?

That’s what I’m trying to figure out
 
  • #10
y90x said:
That’s what I’m trying to figure out
Use the formula you quoted in post #3. The unknown mass of the rock, m, appears in the answer.
 
  • #11
haruspex said:
Use the formula you quoted in post #3. The unknown mass of the rock, m, appears in the answer.

I’m not trying to find mass, the mass is given .
I’m looking for gravitational acceleration,
I keep getting 0.65 but I’m not sure if that’s correct
 
  • #12
y90x said:
I’m not trying to find mass, the mass is given .
No, the mass you are given is the mass of the planet.
In post #5 you complained that you are not given the mass of the rock, so you cannot use the formula to find the force on it. I keep trying to explain to you that that is not a problem. You can find an expression for the force which includes the mass of the rock as an unknown.
y90x said:
I keep getting 0.65 but I’m not sure if that’s correct
0.65 what? m/s2?
Please post all your working.
 
  • #13
haruspex said:
No, the mass you are given is the mass of the planet.
In post #5 you complained that you are not given the mass of the rock, so you cannot use the formula to find the force on it. I keep trying to explain to you that that is not a problem. You can find an expression for the force which includes the mass of the rock as an unknown.

0.65 what? m/s2?
Please post all your working.

Image1513028246.007019.jpg
 

Attachments

  • Image1513028246.007019.jpg
    Image1513028246.007019.jpg
    25 KB · Views: 868

Similar threads

Replies
2
Views
2K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
Replies
15
Views
3K