Find the ratio of maximum height to radius of planet

Click For Summary
SUMMARY

The discussion centers on calculating the ratio of maximum height to radius of a planet when a projectile is launched at 61% of the escape speed. The initial approach incorrectly applies the kinematic equation Vf² = Vi² - 2gh, which is not valid when the height is comparable to the planet's radius. The correct method involves using the law of conservation of energy, leading to the conclusion that the ratio h/R equals 0.3721, derived from the energy equations and gravitational principles.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy principles
  • Familiarity with the concepts of escape velocity and projectile motion
  • Knowledge of the law of conservation of energy in physics
  • Basic mathematical skills for manipulating equations involving gravitational constants
NEXT STEPS
  • Study the law of conservation of energy in gravitational fields
  • Learn about escape velocity calculations for different celestial bodies
  • Explore the differences between gravitational acceleration and universal gravitational constant
  • Review advanced projectile motion equations and their limitations in varying gravitational contexts
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in celestial mechanics and gravitational effects on projectile motion.

Elm956
Messages
1
Reaction score
0

Homework Statement


Hint: Disregard any dissipative effects of the
atmosphere of the planet.

A projectile is launched from the surface of a planet (mass M, radius R)
with a launch speed equal to 61 percent of the escape speed for that
planet. The projectile will rise to a maximum height and fall back to the
surface of the planet. What will be the ratio of its maximum height above
the surface to radius of the planet, h/R?

Homework Equations


1/2mv² = -Gmm/R
Vf² = Vi² - 2gh

The Attempt at a Solution


1. 1/2mv² = -Gmm/R
v² = -2Gm/R
v = .61√(2Gm/R)

2. Vf² = Vi² - 2gh
0 = Vi² - 2gh
Vi² = 2gh
Vi = √2gh

3. .61√(2Gm/R) = √(2Gh)
.7442(Gm/R) = 2(Gmh/R²)
.7442Gm = 2(Gmh/R)
.7442 = 2h/R
h/R = .3721

Any thoughts or advice on the incorrect answer? Thanks in advance.
 
Physics news on Phys.org
You made two serious mistakes:

i. You can not apply the formula Vf² = Vi² - 2gh when the height a projectile reaches is comparable with the radius of the planet.

ii. "g" in the formula above is the free -fall acceleration at the surface of the planet, and entirely different from the universal gravitational constant G.

Use the law of conservation of energy in the gravitational field of the planet.

ehild
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 21 ·
Replies
21
Views
4K
Replies
40
Views
3K
Replies
5
Views
2K
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K