Radial Distance of Projectile Shot from Earth's Surface

In summary, the conversation discusses a projectile being shot away from Earth's surface without considering the Earth's rotation. It asks for the multiple of Earth's radius that gives the radial distance the projectile reaches when its initial speed is 0.758 of the escape speed and its initial kinetic energy is 0.758 of the energy required to escape Earth. The equation for escape velocity is given as v=\sqrt{2GM/R}. For part a, the speed is multiplied by 0.758 and plugged into the equation, while for part b, the entire 1/2mv^2 term is multiplied by 0.758. The use of the standard gravitational parameter (μ) and the difference between initial velocity and initial kinetic energy are
  • #1
mariahkraft
6
0
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.758 of the escape speed from Earth and (b) its initial kinetic energy is 0.758 of the kinetic energy required to escape Earth?

So basically what i got from lecture is that K+U=1/2mv2+-GmM/R=0
v=[tex]\sqrt{2GM/R}[/tex]

for part a would I just multiply v=[tex]\sqrt{2GM/R}[/tex] by .758 then plug that in for v in the 1/2 mv2?

then for part b would i multiply the entire 1/2mv2 by .758? Also one more thing for part b am i using my answer from part a for the speed or am i using standard escape speed?

I have my final on monday and I am trying to review for the test any help would be awesome Thanks!.
 
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  • #2
mariahkraft said:
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.758 of the escape speed from Earth and (b) its initial kinetic energy is 0.758 of the kinetic energy required to escape Earth?

So basically what i got from lecture is that K+U=1/2mv2+-GmM/R=0
v=[tex]\sqrt{2GM/R}[/tex]

for part a would I just multiply v=[tex]\sqrt{2GM/R}[/tex] by .758 then plug that in for v in the 1/2 mv2?

then for part b would i multiply the entire 1/2mv2 by .758? Also one more thing for part b am i using my answer from part a for the speed or am i using standard escape speed?

I have my final on monday and I am trying to review for the test any help would be awesome Thanks!.

Welcome to PF.

Your first equation is for escape velocity. But for escape velocity it is assumed that potential is 0 and kinetic energy is 0.

So what they are really giving you is 1/2m(.758v)2 + (-μ*m/Re) = PE at max height which is given by -μ*m/Rmax

The second part wants you to recognize the difference between initial velocity and initial KE which is v:v2

(Note I used μ as GM the standard gravitational parameter for Earth which numerically is 398K in SI units)
 
  • #3
thank you!
 

FAQ: Radial Distance of Projectile Shot from Earth's Surface

What is the radial distance of a projectile shot from Earth's surface?

The radial distance of a projectile shot from Earth's surface is the distance between the center of the Earth and the point where the projectile lands on the Earth's surface.

How is the radial distance of a projectile shot from Earth's surface measured?

The radial distance can be measured using various methods, such as using a ruler or measuring tape on a map or using geographic coordinates and a GPS device.

How does the angle of the projectile's trajectory affect its radial distance from Earth's surface?

The angle of the projectile's trajectory will affect its radial distance from Earth's surface. A projectile shot at a higher angle will have a longer radial distance than one shot at a lower angle, as it will travel a greater distance from the Earth's surface before landing.

Can the Earth's curvature affect the radial distance of a projectile shot from its surface?

Yes, the Earth's curvature can affect the radial distance of a projectile shot from its surface. As the Earth is curved, the distance between the starting point and the landing point of a projectile will be longer than the straight line distance due to the Earth's curvature.

How does the initial velocity of a projectile affect its radial distance from Earth's surface?

The initial velocity of a projectile will affect its radial distance from Earth's surface. A projectile with a higher initial velocity will travel further and have a longer radial distance than one with a lower initial velocity, assuming all other factors remain constant.

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