# Help with 6 of the 20 universal gravitational law questions that I have due

1. Oct 26, 2008

### DK6021023

1. Planet A has a period of 1 year and an average distance from the sun of 1.63 × 1011 meters. If the average distance from the sun for planet B is 0.53 × 1011 meters, what is its period to the nearest hundredth of a year?

2. Two objects attract each other gravitationally with a force of when they are 0.25 m apart. Their total mass is 4.0 kg. Find their individual masses.

3. Because the Earth rotates once per day, the apparent acceleration of gravity at the equator is slightly less than it would be if the Earth didn’t rotate. Estimate the magnitude of this effect. What fraction of g is this?

4. How long would a day be if the Earth were rotating so fast that objects at the equator were apparently weightless?

5. Jupiter is about 320 times as massive as the Earth. Thus, it has been claimed that a person would be crushed by the force of gravity on a planet the size of Jupiter since people can’t survive more than a few g’s. Calculate the number of g’s a person would experience at the equator of such a planet. Use the following data for Jupiter: equatorial rotation Take the centripetal acceleration into account.

6. Astronomers using the Hubble Space Telescope deduced the presence of an extremely massive core in the distant galaxy M87, so dense that it could be a black hole (from which no light escapes). They did this by measuring the speed of gas clouds orbiting the core to be at a distance of 60 light-years from the core. Deduce the mass of the core, and compare it to the mass of our Sun.

2. Oct 27, 2008

### tiny-tim

Welcome to PF!

Hi DK6021023! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

3. Oct 27, 2008

### DK6021023

Re: Welcome to PF!

I got done everything. Now I only need help with #5. I'm also confused with #3.

Here are my attempts:

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#5.

m = 1.9*10^27
r = 7.1*10^4 km = 7.1*10^7 m
T = 9h55min = 540+55 min = 35700 s

a = [4(pi^2)(7.1*10^7)] / 35700^2
a = 2.199 m/s^2 (I'm assuming this is gravity on Jupiter)
This makes no sense because the gravity is supposed to me way higher than on Earth.

So I tried using the following method:
(F is the net force or the force of gravity)
F = G (m) / r^2
F = (6.67*10^-11)(1.9*10^27)/(7.1*10^7)
F = 1784929577

F = ma
a = F/m
a = (1784929577)/(1.9*10^27)
a = 9.394*10^-19 m/s^2
This blowed my head off, it makes no sense to me whatsoever.

The final answer is supposed to be something close to 39.9u, but i don't know how to get there.

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#3.

a = [4(pi^2)r]/T^2
a = [4(pi^2)(6.38*10^6)]/86400^2
a = 0.03374 m/s^2

so 9.8 m/s^2 + 0.03374 m/s^2 would be the gravity at the equator if Earth didn't rotate.

and 9.8/0.03374 would be the fraction.

but I don't know if I'm right.

4. Oct 27, 2008

### DK6021023

Re: Welcome to PF!

and for #4, you just find the velocity when N=0, and then find the period, right?

5. Oct 28, 2008

### tiny-tim

Hi DK6021023!
No, F = G (Mm) / r^2;

a = G (m) / r^2.

You don't need to divide by m again.

(m is the planet's mass, M is your mass!).

And the centripetal acceleration should be very small compared with the gravity.
Yes, but it's 0.03374/9.8, isn't it?
That's right, but it's a rather complicated way of putting it.

Objects are weightless when the gravity equals the c. a.