What is the missile's speed at the peak of its trajectory?

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Homework Help Overview

The problem involves determining the speed of a missile at the peak of its trajectory, given its maximum altitude and launch speed. The subject area includes concepts from kinematics and gravitational potential energy.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic and potential energy, questioning whether the speed at the peak is zero and exploring the implications of altitude versus distance from the Earth's center.

Discussion Status

Participants are actively engaging with the problem, checking calculations and clarifying the correct distance to the center of the Earth. There is a focus on ensuring proper unit conversion and understanding the physical setup of the problem.

Contextual Notes

There is an emphasis on the need to correctly account for the distance from the center of the Earth, which includes both the altitude of the missile and the radius of the Earth. Participants are navigating through potential misunderstandings regarding these measurements.

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Homework Statement



A missile's trajectory takes it to a maximum altitude of 1200 km.
If its launch speed is 6.2 km/s , how fast is it moving at the peak of its trajectory?

Homework Equations



K+U = K_o+U_o

U(r) = -GMm/r

The Attempt at a Solution



Alright some people have been telling me its zero. Intuitively that's what I thought, but
I guess its wrong. Nevertheless, here is my attempt :

K+U = K_o +U_o;

=

1/2mV^2 - GM_eM/r_d = 1/2mV_o^2 - GM_em / r_e

and solving for
V^2 = V_o^2 - 2GM_e/r_e + GM_e/r_d

where,
V_o = 6.2km -> 6.2*1000m
G = universal graity = 6.67*10^-11
M_e = mass of Earth = 5.98*10^24
r_e = radius of Earth = 6.37 * 10^6;
r_d = distance between the object and the center of earth
= 1200km -> 1200*1000m

and my answer is terms of meters
~2400m

so dividing by 1000

and

v_p ~ 24m/s
 
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tnutty said:
Alright some people have been telling me its zero. Intuitively that's what I thought, but
I guess its wrong.
Why would you think that? (It may be moving horizontally at the peak.)

Nevertheless, here is my attempt :

K+U = K_o +U_o;

=

1/2mV^2 - GM_eM/r_d = 1/2mV_o^2 - GM_em / r_e
Looks good.

and solving for
V^2 = V_o^2 - 2GM_e/r_e + GM_e/r_d
You left off a factor of 2 in that last term.

where,
V_o = 6.2km -> 6.2*1000m
G = universal graity = 6.67*10^-11
M_e = mass of Earth = 5.98*10^24
r_e = radius of Earth = 6.37 * 10^6;
r_d = distance between the object and the center of earth
= 1200km -> 1200*1000m
1200km is just the altitude, not the distance to the center of the earth.
 


so is this correct?

Although my formula missed a factor of 2, I still did the calculation with the 2 factored in.
Can you check my calculations?
 


Did you correct your value of r_d?
 


no, I am not sure why I should. The units won't match.

I converted everything into meter and the final answer into km
 


The issue is not units, but that you are using r_d = 1200 km. That's the altitude; you need the distance to the center of the earth.
 


so is the distance 1200km + Earth's radius?
 


how do I get the distance to the center of the Earth of the object?
 


tnutty said:
so is the distance 1200km + Earth's radius?
Yes, that's r_d.
tnutty said:
how do I get the distance to the center of the Earth of the object?
See above.
 
  • #10


So the distance is 1200 * 1000m + Earth's radius
because
1) It they are in both meters.

2) Because the projectory is launched from the Earth's surface, or top of the earth?
 
  • #11


Good
 
  • #12


Cool, thanks!