Initial speed of an object launched from Mars

Click For Summary
The discussion focuses on calculating the initial speed of an object launched from Mars using gravitational potential energy and kinetic energy equations. The user initially miscalculated the initial velocity due to incorrect unit conversion of the radius from kilometers to meters. After correcting the radius, the calculation yielded a revised initial speed of approximately 153,776.815 m/s. The importance of unit consistency in physics calculations is emphasized. The user expresses gratitude for the assistance in identifying the mistake.
marjine
Messages
10
Reaction score
1
Homework Statement
The radius of Mars (from the center of just above the atmosphere) is 3400 km, and its mass is 6 × 1023 kg. An object is launched straight up from just above the atmosphere of Mars. What initial speed (vi) is needed so that when the object is far from Mars, its final speed (vf) is 500 m/s?
Relevant Equations
G = 6.7E-11
Conservation of energy principle: U1 + K1 = U2 + K2
U1 = -GMm/r
K1 = (1/2)mvi^2
U2 = as r approaches infinity, U2 approaches zero
K2 = (1/2)mvf^2

(1/2)mvi^2 - GMm/r = (1/2)mvf^2 + 0

vi = √(vf^2 + (2GMm)/r) = √(250,000 + 2(6.7 E-11)(6 E23)/3400) = 153776.815

But that is not the correct answer, can anybody see my mistake/misunderstanding?
 
Physics news on Phys.org
The radius (r) is given in km, isn't it?
 
  • Like
Likes berkeman and PeroK
nasu said:
The radius (r) is given in km, isn't it?
omg... converted it to m and fixed it, thank you so much
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
1K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 3 ·
Replies
3
Views
4K