Change in speed needed for a satellite to graze the surface

[Flipmeister]

Homework Statement

A moon lander is orbiting the moon at an altitude of 1000km. By what percentage must it decrease its speed so as to graze the moon's surface one-half period later?

Homework Equations

$$F_{G}=\frac{GMm}{R^2}\\ T=√(\frac{4\pi^2 R^3}{GM})=\frac{2\pi R}{v}\\ v=\frac{2\pi R}{T}\\ KE_{rot}=\frac{1}{2}Iω^2\\ KE_{trans}=\frac{1}{2}mv^2\\ U_g=-\frac{GMm}{r}\\ Δs=v_i t+\frac{1}{2}a t^2\\$$
Radius of the moon is 1.74x10^6 m, so the radius of the 1000km orbit is 2.74x10^6 m.

The Attempt at a Solution

I can easily calculate the velocity it had in its original 1000km orbit: ~1339 m/s. So I just need to find the new velocity, which should be a little under 1339 m/s...

Since I need to find a speed to get from one distance from the center of the moon to another in half a period, I tried starting off with kinematics. Seeing as there is no tangential acceleration, only radial, the tangential displacement, which is 0.5 of the circumference, should only depend on the initial tangential velocity (which is the initial velocity, since it was in orbit at first) and time. So I tried to find the time it takes to go from the initial to final radius, given by R(final) - R(initial) = v(radial)*t + 0.5at^2, so
$$10^6 m = \frac{1}{2}at^2$$
Since the initial v(radial) = 0. Now the poop hits the fan. a=GM/R^2 so I get t=R(6.38x10^(-4)); already looking ugly. I end up with an velocity of (1.35x10^10 m^2/s)/R. The R there makes no sense, since that was the R from the acceleration and varies with time. I'm guessing the problem is the a I am using in my kinematic equation... a depends on R, which frankly seems to depend on a. I feel there is calculus involved; do I integrate/differentiate something? How do I use a in the equation so I can find my time?

Am I supposed to be using conservation of energy? I've considered it as well, but I couldn't get an initial velocity there because I have no final velocity to work with as far as I can see.

 

[gneill]

Homework Statement

A moon lander is orbiting the moon at an altitude of 1000km. By what percentage must it decrease its speed so as to graze the moon's surface one-half period later?

Homework Equations

$$F_{G}=\frac{GMm}{R^2}\\ T=√(\frac{4\pi^2 R^3}{GM})=\frac{2\pi R}{v}\\ v=\frac{2\pi R}{T}\\ KE_{rot}=\frac{1}{2}Iω^2\\ KE_{trans}=\frac{1}{2}mv^2\\ U_g=-\frac{GMm}{r}\\ Δs=v_i t+\frac{1}{2}a t^2\\$$
Radius of the moon is 1.74x10^6 m, so the radius of the 1000km orbit is 2.74x10^6 m.

The Attempt at a Solution

I can easily calculate the velocity it had in its original 1000km orbit: ~1339 m/s. So I just need to find the new velocity, which should be a little under 1339 m/s...

Since I need to find a speed to get from one distance from the center of the moon to another in half a period, I tried starting off with kinematics. Seeing as there is no tangential acceleration, only radial, the tangential displacement, which is 0.5 of the circumference, should only depend on the initial tangential velocity (which is the initial velocity, since it was in orbit at first) and time. So I tried to find the time it takes to go from the initial to final radius, given by R(final) - R(initial) = v(radial)*t + 0.5at^2, so
$$10^6 m = \frac{1}{2}at^2$$
Since the initial v(radial) = 0. Now the poop hits the fan. a=GM/R^2 so I get t=R(6.38x10^(-4)); already looking ugly. I end up with an velocity of (1.35x10^10 m^2/s)/R. The R there makes no sense, since that was the R from the acceleration and varies with time. I'm guessing the problem is the a I am using in my kinematic equation... a depends on R, which frankly seems to depend on a. I feel there is calculus involved; do I integrate/differentiate something? How do I use a in the equation so I can find my time?

Am I supposed to be using conservation of energy? I've considered it as well, but I couldn't get an initial velocity there because I have no final velocity to work with as far as I can see.

Hi Flipmeister, welcome to Physics Forums.

Think in terms of changing the orbit from circular to elliptical such that the aposelene is at the starting distance and periselene is at the surface. In other words, much like a Hohmann transfer trajectory.

 

[Flipmeister]

Hi Flipmeister, welcome to Physics Forums.

Think in terms of changing the orbit from circular to elliptical such that the aposelene is at the starting distance and periselene is at the surface. In other words, much like a Hohmann transfer trajectory.

Thanks! My textbook hardly touched on the subject of Hohmann transfer orbit, so I was still lost for a while until I did some googling. :tongue: