Question

P1:Suppose an object is launched from Earth with 0.52 times the
escape speed. How many multiples of Earth's radius (R_{E} =
6.37 x 10^{6} m) in radial distance will the object reach
before falling back toward Earth? The distances are measured
relative to Earth's center, so a ratio of 1.00 would correspond to
an object on Earth's surface. For this problem, neglect Earth's
rotation and the effect of its atmosphere.

For reference, Earth's mass is 5.972 x 10^{24} kg.

Your answer is a ratio and thus unitless. Enter only a number in the answer box.

P2

Suppose an object is launched from Earth with 0.6 times the
kinetic energy for escape. How many multiples of Earth's radius
(R_{E} = 6.37 x 10^{6} m) in radial distance will
the object reach before falling back toward Earth? The distances
are measured relative to Earth's center, so a ratio of 1.00 would
correspond to an object on Earth's surface. For this problem,
neglect Earth's rotation and the effect of its atmosphere.

For reference, Earth's mass is 5.972 x 10^{24} kg.

Your answer is a ratio and thus unitless. Enter only a number in the answer box.

Answer #1

**P1)**

**we know, Escape speed on the surface of the Earth, ve =
sqrt(2*G*Me/Re)**

**let r is the distance from center of the earth to the
maximum height of the object reached.**

**let m is the mass of the object.**

**and v = 0.52*ve**

**now apply conservation of energy**

**-G*Me*m/Re + (1/2)*m*v^2 = -G*Me*m/r**

**-G*Me*m/Re + (1/2)*m*0.52^2*(2*G*Me/Re) =
-G*Me*m/r**

**-G*Me*m/Re*(1 - 0.52^2) = -G*Me*m/r**

**0.7296/Re = 1/r**

**r = Re/0.7296**

**= 1.37*Re
<<<<<<<<-------------------Answer**

**P2)**

**Again apply conservation of energy**

**-G*Me*m/Re + (1/2)*m*v^2 = -G*Me*m/r**

**-G*Me*m/Re + 0.6*(1/2)*m*ve^2 = -G*Me*m/r**

**-G*Me*m/Re + 0.6*(1/2)*m*(2*G*Me/Re) =
-G*Me*m/r**

**-G*Me*m/Re*(1 - 0.6) = -G*Me*m/r**

**0.4/Re = 1/r**

**r = Re/0.4**

**= 2.50*Re
<<<<<<<<-------------------Answer**

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