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 10⁶ 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 10²⁴ 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 10⁶ 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 10²⁴ 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