Question

A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius of mars = 3395 km; G = 6.67× 10-11 N m2 kg–2. Question8.25 Gravitation

Answer 1

Initial K.E. of rocket = 1/2mv²

Initial P.E of rocket = -GMm/R

Total enrgy = 1/2 mv ² - GMm/R

ATQ , 

80/100 x 1/2mv² - GMm/R 

0.4mv² - GMm/R

Let the maximum height reached by rocket = "h"

At final position the KE = 0

Energy of rocket at "h"

= -GMm(R+h)

Now , By law of conservation of energy 

0.4 mv² - GMm/R = -GMm/(R+h)

0.4v² = GM/R - GM(R+h)

         = GM(1/r - 1/(R+h))
          
         = GM( R+h -R/ R(R+h) ) 

         = GMh /R(R+h) 

or 

R+h/h = GM/0.4v²R 

R/h + 1 = GM/0.4v²R 

R/h = (GM / 0.4v²R) - 1 

h = ( R/GM) / 0.4v²R - 1 

h = 0.4 R²v²/ GM - 0.4 v²R 

h = 0.4 x (3.395 x 10⁶) ² x (2 x 10³) ²/ 6.67 x 10¹¹ x 6.4 x 10²³ - 0.4 x (2 x 10)² x (3.395 x 10⁶) 

h = 495 x 10³m = 495 km 

Answer 2

Given,
speed of the rocket ( v) = 2 km/s
mass of the Mars ( Mm) = 6.4 × 10²³ Kg
radius of the Mars (RM ) = 3395 km = 3.395 × 10^6 m

inital K.E of the rocket = 1/2 mv²
due to Martian resistance 20% of KE of the rocket is lost .
so, available K.E = 80% of 1/2mv²
= 2/5 mv²--------(1)

Let the rocket be reach at height h from the surface of the Mars .
so, increase PE ° = PEf - PEi
= -GMm.m/(Rm + h) - [ - GMm.m/Rm]
= -GMm.m × h/Rm.(Rm + h)

now, according to Law of conservation of energy .
increase in PE = available total KE
GMm.m×h/Rm(Rm+h) = 2/5 mv²
GMm/Rm = 2/5 v².(Rm + h)/h
5GMm/2Rm.v² = Rm/h + 1
after putting values of G, Mm, Rm and v
5×6.67×10^-11×6.4×10²³/2×3.395×10^6×(2×10³)² = Rm/h +1

7.85 = Rm/h +1
Rm/h = 6.85
h = Rm/6.85 = 3395/6.85 ≈ 495 km