What Is the Speed of a Spacecraft Near Mars?

[bibrahim6]

A spacecraft is coasting toward Mars. The mass of Mars is 6.4 × 1023 kg and its radius is 3400 km (3.4 × 106 m). When the spacecraft is 8500 km (8.5 × 106 m) from the center of Mars, the spacecraft 's speed is 2250 m/s. Later, when the spacecraft is 5300 km (5.3 × 106 m) from the center of Mars, what is its speed? Assume that the effects of Mars's two tiny moons, the other planets, and the Sun are negligible. Precision is required to land on Mars, so make an accurate calculation, not a rough, approximate calculation.

I know the answer is 3340 m/s but i can't seem to get it.
Any help will do

 

[nil1996]

Use work energy theorem.The change in potential energy is equal to change in kinetic energy.
Welcome to forum

 

[gneill]

A spacecraft is coasting toward Mars. The mass of Mars is 6.4 × 1023 kg and its radius is 3400 km (3.4 × 106 m). When the spacecraft is 8500 km (8.5 × 106 m) from the center of Mars, the spacecraft 's speed is 2250 m/s. Later, when the spacecraft is 5300 km (5.3 × 106 m) from the center of Mars, what is its speed? Assume that the effects of Mars's two tiny moons, the other planets, and the Sun are negligible. Precision is required to land on Mars, so make an accurate calculation, not a rough, approximate calculation.

I know the answer is 3340 m/s but i can't seem to get it.
Any help will do

What have you tried? What equations are relevant (the posting template is there for a reason!). What concepts are involved? We can't help if we can't see where you need help!

 

[bibrahim6]

Ive tried delta K = delta U so 1/2 m(vf-vi)^2 = -G (Mm/(rf-ri)) but i get initial is equal to final since the potential is very small

 

[gneill]

Let's see some numbers. What's the 'per kg' change in potential energy due to moving between the two positions?

 

[gneill]

Ive tried delta K = delta U so 1/2 m(vf-vi)^2 = -G (Mm/(rf-ri)) but i get initial is equal to final since the potential is very small

Also note that (vf - vi)² is not the same as vf² - vi². The change in kinetic energy is given by $\frac{1}{2}m (vf^2 - vi^2)$.

 

[bibrahim6]

So this was my attempt

(.5)m( (vf)^2 - (2250)^2 )= -(6.67e-11)* ( (m *6.4 × 1023 ) / (5.3e6 - 8.3e6) )

so m= mass of spacecraft I'm assuming it cancels out and i end up getting
vf= 2250 which is not correct

and does the radius of Mars come into play in this problem?

 

[gneill]

So this was my attempt

(.5)m( (vf)^2 - (2250)^2 )= -(6.67e-11)* ( (m *6.4 × 1023 ) / (5.3e6 - 8.3e6) )

so m= mass of spacecraft I'm assuming it cancels out and i end up getting
vf= 2250 which is not correct

and does the radius of Mars come into play in this problem?

Your algebra is causing you problems. For the PE change,

$$\frac{GMm}{r_f} - \frac{GMm}{r_i} \neq \frac{GMm}{r_f - r_i}$$

The radius of Mars won't come into play, and yes the mass of the spacecraft will cancel out.

 

[bibrahim6]

I corrected that and ended up getting 3299 m/s...can you lay out an exact equation please I am just not seeing my error

 

[gneill]

I corrected that and ended up getting 3299 m/s...can you lay out an exact equation please I am just not seeing my error

I can't give you the equation, but I can point out errors you make. Why don't you write out the conservation of energy equation in symbols and I'll verify it for you. Then we can look at the numbers.

 

[bibrahim6]

Nevermind I put 8.3 instead of 8.5 THANK YOU FOR THE HELP!

 

[gneill]

No problem. Glad to help.