Newton's Law of Universal Gravitation Problem

[ova5676]

Homework Statement

Answers: 56. 6.16 x 10^17 N, 57. 894 N

Homework Equations

• Fg = G*m1*m2/r^2
• g = G*M/r^2

The Attempt at a Solution

56. Since the closest planet to Z is Y I figured the gravitational force could be calculated by just finding the gravitational force between Z and Y, which I did, but it didn't turn out to be the answer. So I'm thinking X is involved somehow, but I have no idea, I didn't really learn this.

57. This, you'd think, would be straight forward. I could just use the first equation using G, m1 (100kg), m2 (the mass of the Earth - 5.9742 x 10^24 kg) and the radius 300 km, but that wasn't the answer either? I'm stuck!

 

[Idividebyzero]

hmm, what about sumation of the forces on planet z both planet x and planet y have a gravitational force acting in accordance with z. this should be correct imho because the question asks for the NET gravitational force just one would be a component of the net force. you would have to use the law for the first distance added to the law for the combined distances... ie

Fnet=(G*m1*m2/r1^2) + (G*m1*m2/(r1+r2)^2)

 

[Idividebyzero]

what units is it asking for the force of gravity? i normally convert all quantities to SI units where a Newton N=Kg*m/s/s. here we have kg for the mass but the r value is in km, i normally would convert to meters as customary to myself... if its asking for Newtons you might try converting to meters and calculating again...

 

[gneill]

The gravitational force does not depend upon which bodies are closest. Every mass interacts with every other mas according to the inverse square law. So for your question 56, sum the forces due to both X and Y according to their respective distances.

For 57, consider what the R represents in the gravitational force equation, then re-read the question statement.

 

[ova5676]

The gravitational force does not depend upon which bodies are closest. Every mass interacts with every other mas according to the inverse square law. So for your question 56, sum the forces due to both X and Y according to their respective distances.

For 57, consider what the R represents in the gravitational force equation, then re-read the question statement.

It's kind of late, so excuse my mind-numbness here but what do you mean by sum of the forces due to both X and Y? Do you mean

Fnet = Fg between X and Y + Fg between Y and Z? or just Fg between X and Y?

And for 57, r means radius, which is 300 km above the Earth's surface, isn't that the distance = radius? What do you mean? I don't get it?

 

[SammyS]

It's kind of late, so excuse my mind-numbness here but what do you mean by sum of the forces due to both X and Y? Do you mean

Fnet = Fg between X and Y + Fg between Y and Z? or just Fg between X and Y?

Planet Y exerts a force on planet Z AND planet X exerts a force on planet Z. Find the sum of these forces.

And for 57, r means radius, which is 300 km above the Earth's surface, isn't that the distance = radius? What do you mean? I don't get it?

The force depends upon the astronaut's distance from the Earth's center!

 

[ova5676]

Planet Y exerts a force on planet Z AND planet X exerts a force on planet Z. Find the sum of these forces.


The force depends upon the astronaut's distance from the Earth's center!

So if that's not the radius, then what do I use?

 

[gneill]

So if that's not the radius, then what do I use?

If it's not the radial distance, then calculate the radial distance! Height above Earth's surface is not the same as the distance from the Earth's center!

 

[ova5676]

If it's not the radial distance, then calculate the radial distance! Height above Earth's surface is not the same as the distance from the Earth's center!

I understand, but how would I get radial distance?

 

[gneill]

Look up a value for the Earth's radius. That's the distance from the Earth's center to the Earth's surface. In the problem you're given the distance of the astronaut from the Earth's surface. Do the addition.