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