Question

A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, and the radius of Planet X is estimated to be twice that of Earth. The estimated mass and radius of Planet X are used to calculate the minimum escape speed, vc , for an object launched from the surface of the planet. If the actual mass and/or radius of the planet are slightly different from the estimated values.
Required:
How will the actual escape speed va for the surface of Planet X compare to vc?

Answer 1

Answer:

    vₐ = v_c  $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

          Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

         Em_f = U = - G Mm / R₂

energy is conserved

           Em₀ = Em_f

           ½ m v_c² - G Mm / R = - G Mm / R₂

           v_c² = 2 G M (1 /R -  1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

           v_c = $\sqrt{\frac{2GM}{R} }$

now indicates that the mass and radius of the planet changes slightly

            M ’= M + ΔM = M ( $1+ \frac{\Delta M}{M}$ )

            R ’= R + ΔR = R ( $1 + \frac{\Delta R}{R}$ )

we substitute

           vₐ = $\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }$

         

let's use a serial expansion

           √(1 ±x) = 1 ± ½ x +…

we substitute

         vₐ = v_ c ( $(1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )$)

we make the product and keep the terms linear

        vₐ = v_c  $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$