Physics Question......many thanks
posted by Jessy .
A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.
(a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is onefifth of the escape speed from Earth?
____ times RE
(b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is onefourth of the kinetic energy required to escape Earth?
_____ timesRE
(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?
These problems all can be solved by using the Conservation of Energy relationship
Total Energy = KE + PE =
(1/2) m V^2  GMm/R = constant
wher M is the mass of the earth, m is the mass of the projectile and R is the distance from the center of the Earth.
The "escape speed" Ve is the launch velocity that allows V to be zero when R = infinity. Thus
(1/2) m Ve^2  G M m/Re = constant
Ve = sqrt (2 G M/Re)
Here is how to do (a):
If V = (1/5) Ve at R = Re, then
(1/2) m V^2  G M m/R = constant
= (1/50) m Ve^2  G M m/Re
When V = 0 (maximum value of R),
(1/50) Ve^2 = GM [(1/Re  (1/R)]
(1/25) G M/Re = G M [(1/Re  (1/R)]
GM/R = (24/25)(GM/Re)
R/Re = 25/24
Proceeds imilarly for (b)
For (c), the least mechanical energy (including potential energy), is zero, since potential energy is defined as zero at infinite distance, as is almost always done in inversesquare law situations. This is the case when PE =  G M m/R
Your explanaion is very clear to me. Thank you.

Sorry, your explanation made too many logical leaps. How did you get from 1/5 to 1/50??? It's garbage like this that makes you want to never see physics problems again.

Amen Max!

1/2(1/5Ve)^2 = 1/2 * 1/25 * Ve^2 = 1/50Ve^2

why is it 1/50??

Because (1/5Ve)^2 = 1/25Ve^2 You then multiply the 1/25 by the 1/2 to get 1/50.
Respond to this Question
Similar Questions

Gravitation help..........thanks
A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth. (a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is onefifth of the escape speed … 
physics
1. (a) How far below an initial straightline path will a projectile fall in eight seconds? 
physics
A 3Kg projectile is fired straight upward from earth's surface with a speed that is onefourth the escape speed. Neglecting air resistance. a) What is the Kinetic energy of this projectile when it is launched? 
physics
A projectile with mass of 139 kg is launched straight up from the Earth's surface with an initial speed vi. What magnitude of vi enables the projectile to just reach a maximum height of 5.8RE, measured from the center of the Earth? 
PHYSICS
At the Earth's surface, a projectile is launched straight up at a speed of 8.4 km/s. To what height will it rise? 
Physics
PROJECTILE LAUNCHED FROM EARTH'S SURFACE? 
Physics
Neglect the gravity of the Moon, neglect atmospheric friction, and neglect the rotational velocity of the Earth in the following problem. A long time ago, Jules Verne, in his book From Earth to the Moon (1865), suggested sending an … 
Physics
A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface of Earth will it go? 
physics
A spherical nonrotating planet (with no atmosphere) has mass 4 kg and radius 9000 km. A projectile of mass is fired from the surface of the planet at a point with a speed at an angle with respect to the radial direction. In its subsequent … 
Physics
At the Earth's surface, a projectile is launched straight up at a speed of 9.9 km/s. To what height will it rise?