MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
WE_W08D2-1 Worked Example Gravitational Slingshot Solution
A spacecraft of mass m1 with a speed v1,i approaches Saturn which is moving in the opposite
direction with a speed vs . After interacting gravitationally with Saturn, the spacecraft swings
around Saturn and heads off in the opposite direction it approached. The mass of Saturn is ms .
Find the final speed of the spacecraft after it is far enough away from Saturn to be nearly free of
Saturn’s gravitational pull.
Solution:
!
Choose the positive x-direction such that the initial velocity of the spacecraft is v1,i = v1,i î , and
!
the final velocity is v1, f = !v1, f î . The change in the velocity of Saturn is negligible due to the
!
!
interaction so v s,i = v s, f = !vs î . Our two equations corresponding to conservation of momentum
and energy are (see appendix for derivation)
m1vx,1,0 + ms vx,s,i = m1vx,1, f + ms vx,s, f .
(1)
vx,1,i + vx,1, f = vx,s,i + vx,s, f .
(2)
We are assuming that vx,s,i ! vx,s, f = !vs . Therefore Eq. (2) becomes
vx,1,i + vx,1, f = !2vs .
(3)
Because we chose our coordinate system such that vx,1, f = !v1, f and vx,1,i = v1,i , we can solve
equation (3) for the final speed of the spacecraft and determine that
v1, f = +2vs + v1,i .
(4)
Appendix: Consider a one-dimensional completely elastic collision between two objects moving
in the x direction. One object, with mass m1 and initial x-component of the velocity vx,1,i ,
collides with an object of mass m2 and initial x-component of the velocity vx,2,i . The scalar
components vx1,0 and vx 2,0 can be positive, negative or zero. No forces other than the interaction
force between the objects act during the collision. After the collision, the velocity components
are vx,1, f and vx,2, f .
Equating the momentum components before and after the collision gives the relation
m1vx,1,i + m2 vx,2,i = m1vx,1, f + m2 vx,2, f .
(5)
Equating the kinetic energy before and after the collision gives the relation
2
2
2
2
m1vx,1,i
+ m2 vx,2,i
= m1vx,1,
+ m2 vx,2,
f
f
(6)
where the factor of 1/ 2 before each term in Equation (6) has been suppressed. Rewrite these
equations as
m1 (vx,1,i ! vx,1, f ) = m2 (vx,2, f ! vx,2,i )
(7)
2
2
2
2
m1 (vx,1,i
! vx,1,
) = m2 (vx,2,
! vx,2,i
)
f
f
(8)
The second equation above can be written as
m1 (vx,1,i ! vx,1, f )(vx,1,i + vx,1, f ) = m2 (vx,2, f ! vx,2,i )(vx,2, f + vx,2,i ).
(9)
Divide the kinetic energy equation by the momentum equation, yielding
vx.1,i + vx,1, f = vx,2,i + vx,2, f .
(10)