Question

where would the spaceprobe experience the strongest net (or total) gravitional force exerted on it by Earth and Mars

Answer 1

Answer:

r = 41.1 10⁹ m

Explanation:

For this exercise we use the equilibrium condition, that is, we look for the point where the forces are equal  

                  ∑ F = 0

                  F (Earth- probe) - F (Mars- probe) = 0

                  F (Earth- probe) = F (Mars- probe)

Let's use the equation of universal grace, let's measure the distance from the earth, to have a reference system

the distance from Earth to the probe is        R (Earth-probe) = r

the distance from Mars to the probe is        R (Mars -probe) = D - r

where D is the distance between Earth and Mars

                   

                 $G \ \frac{m \ M_{Earth}}{r^2} = G \ \frac{m \ M_{Mars}}{(D-r)^2}$

                 M_earth (D-r)² = M_Mars r²

                 (D-r) = $\sqrt{ \frac{M_{Mars}}{ M_{Earth}} }$    r

                  r ( $1 + \sqrt{ \frac{M_{Mars}}{M_{Earth}} }$) = D

                  r = $\frac{D}{ 1+ \sqrt{\frac{M_{Mars}}{ M_{Earth}} } }$

We look for the values ​​in tables

                  D = 54.6 10⁹ m (minimum)

                  M_earth = 5.98 10²⁴ kg

                  M_Marte = 6.42 10²³ kg = 0.642 10²⁴ kg

                   

let's calculate

                  r = 54.6 10⁹ / (1 + √(0.642/5.98)  )

                  r = 41.1 10⁹ m