2. To leave the gravitational field of a planet, a space rocket must be launched at an escape velocity of:

ve =r2GM r , where G is the gravitational constant G ¡Ö 6.8¡Á10−11Nm2/kg2, M is the mass of the planet and r is its radius.

Suppose that we want to land a space probe on one of a collection of moons of different sizes whose masses are given by M = 1022 + r4 kg. What is the optimal radius to select in order to minimise the escape velocity needed when the space probe departs from the chosen moon?

sorry, to correct the question.

Ve= root 2GM/r

Where G is the gravitational constant 6.8x10^-11,
M is 10^22 + r^4

if possible, please can you provide a step by step answer to the question

v = sqrt [ 2 G M/r ]

v = [ 13.6 *10^-11 (10^22+r^4)/r]^.5

form is

v = [ k (a+r^4)/r ]^.5
we want where dv/dr = 0

0 =dv/dr = .5 k^.5[ (a+r^4)/r ]^-.5 * [r (3 r^3)- (a+r^4)] /r^2

luckily that is zero when
[r (3 r^3)- (a+r^4)] = 0
3 r^4 - r^4 = a
2 r^4 = a
r^4 = a/2
but a = 10^22
so
r^4 = .5 * 10^22 = 50 * 10^20
r = 2.66 * 10^5

CHECK THAT ARITHMETIC !

I think you can get the picture from what I did but I did it fast.

To find the optimal radius that minimizes the escape velocity, we need to differentiate the escape velocity equation with respect to the radius and set it to zero. This will help us find the radius at which the escape velocity is minimized.

First, let's express the equation for the escape velocity by substituting the given mass expression M = 10^22 + r^4 into the original equation:

ve = √(2G(M))/r

ve = √(2G((10^22) + r^4))/r

Now, differentiate the equation with respect to r:

d(ve)/dr = (1/2) * √(2G) * (d(M)/dr) * (r^-1/2) - (1/2) * √(2G) * (M/r^2)

To find the minimum, set the derivative equal to zero:

0 = (1/2) * √(2G) * (d(M)/dr) * (r^-1/2) - (1/2) * √(2G) * (M/r^2)

Rearrange the equation:

(d(M)/dr) * (r^-1/2) = (M/r^2)

Now, let's substitute the expression for M:

(d(10^22 + r^4)/dr) * (r^-1/2) = ((10^22 + r^4)/r^2)

Differentiate M = 10^22 + r^4 with respect to r:

d(10^22 + r^4)/dr = 4r^3

Substitute this result back into the equation:

4r^3 * (r^-1/2) = ((10^22 + r^4)/r^2)

Simplify the equation:

4r^(5/2) = (10^22 + r^4)/r^2

Multiply through by r^2:

4r^(9/2) = 10^22 + r^4

Rearrange the equation:

r^4 - 4r^(9/2) + 10^22 = 0

Unfortunately, we cannot solve this equation analytically. We need to use numerical methods or graphical techniques to find the approximate value of r that satisfies this equation and gives us the optimal radius to minimize the escape velocity.