Calculating Planetary Acceleration in the Solar System | Venus, Mercury Data

In summary, the conversation is about a student struggling with finding the acceleration of planets in the solar system using the equation f=ma. They are using the given mass, distance, and gravitational force of each planet, but are getting numbers that seem too high. Another person suggests using the formula (v^2)/R, where v is velocity and R is the radius of the planet's orbit, which leads to a more accurate result. The student then tries using a different formula (G.m1.m2)/r^2 and gets a better answer.
  • #1
aidan
Evening all.
My teacher has given us the mass, distance, and gravitational force of every planet in the solar system. we have to work out every planets acceleration, speed, and orbital period.
I'm having some trouble with the acceleration. This sshould be a very small number right?
I'm using the equation f=ma. (This is for Venus)
f=ma
a=f/m

f=1.12*10^34
m=4.89*10^24

now when i do f/m i get 2,290,388,548. Thats a huge number and I'm sure that Venus can't be accelerating that much.
My answer for Mercury was 0.0246 m/s.

Please help!
 
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  • #2
Your value for f is way too high. How did you arrive at this number?
 
  • #3
I think the formula you want for the acceleration is (v^2)/R where v is velocity and R is the radius of the planets orbit. Hope that helps.
 
  • #4
Originally posted by S.P.P
I think the formula you want for the acceleration is (v^2)/R where v is velocity and R is the radius of the planets orbit. Hope that helps.

There's nothing wrong with his method, just the starting numbers.

For instance, with your method, he would first have to get the planet's velocity.

given that centripetal force = Mv²/r,

v= [squ] fr/M) =

(using the numbers given)

[squ](1.12 e34 * 1.082e11/4.89e24 = 1.57e 10 m/s

plug this into v²/r gives

(1.57e10)²/1.082e11 = 2286154928

Which is of the same order as the answer he already has.
 
  • #5
i got f doing this:

f = G.m1.m2 / r^2

G = 6.67*10^-11
m1 = 1.99*10^30
m2 = 4.89*10^24
r = 1.08*10^11

which equals, = ...5.56*10^22

is this better??
 
  • #6
Originally posted by aidan
i got f doing this:

f = G.m1.m2 / r^2

G = 6.67*10^-11
m1 = 1.99*10^30
m2 = 4.89*10^24
r = 1.08*10^11

which equals, = ...5.56*10^22

is this better??

Yes.
 

1. What is planetary acceleration?

Planetary acceleration is the rate at which a planet's velocity changes over time due to the pull of gravity from the sun and other celestial bodies.

2. How is planetary acceleration calculated?

To calculate planetary acceleration, we use Newton's second law of motion, which states that force equals mass times acceleration. In this case, the force is the gravitational force between the planet and the sun, and the mass is the planet's mass. By plugging in these values and solving for acceleration, we can calculate the planet's acceleration.

3. What is the significance of calculating planetary acceleration?

Calculating planetary acceleration allows us to better understand the dynamics of our solar system and how planets interact with each other. It also helps us make predictions about the future movement of planets and their orbits.

4. What data is needed to calculate planetary acceleration for Venus and Mercury?

To calculate the acceleration for Venus and Mercury, we need to know the mass of each planet, the distance between the planet and the sun, and the mass of the sun. This data can be found through observations and measurements made by spacecrafts and telescopes.

5. How does planetary acceleration differ between Venus and Mercury?

The planetary acceleration for Venus and Mercury will differ due to the difference in their masses and distances from the sun. Mercury, being closer to the sun, will experience a stronger pull of gravity and therefore have a higher acceleration compared to Venus.

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