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

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SUMMARY

The missile's speed at the peak of its trajectory, calculated using the conservation of mechanical energy, is approximately 24 m/s. The initial launch speed is 6.2 km/s, and the maximum altitude reached is 1200 km. The correct distance to the center of the Earth (r_d) must include the Earth's radius, totaling 1200 km plus 6371 km. The calculations involve gravitational potential energy and kinetic energy, utilizing the formula K + U = K_o + U_o.

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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!