Question

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

Answer 1

r = 41.1 10⁹ m

Step-by-step 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 \ (m \ M_(Earth))/(r^2) = G \ (m \ M_(Mars))/((D-r)^2)`

M_earth (D-r)² = M_Mars r²

(D-r) =
`\sqrt{ (M_(Mars))/( M_(Earth)) }` r

r (
`1 + \sqrt{ (M_(Mars))/(M_(Earth)) }`) = D

r =
`\frac{D}{ 1+ \sqrt{(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