Help with two problems relating to universal gravitation

[TextClick]

Homework Statement

5) Jupiter has a mass about 300 times that of Earth, and its radius is about 11 times that of Earth. What would would be the approximate weight of a 5 kg rock on the surface of Jupiter?

6) A satellite is placed in a circular orbit 100.0 km above Earth's surface. a) What is the speed of the satellite? b) How many minutes does it take the satellite to complete one orbit?

Relevant Equations

Newton's Law of Universal Gravitational Force
F=(Gm1m2/r^2)

Kepler's Third Law
(Ta/Tb)^2=(ra/rb)^3

Newton's Form of Kepler's Third Law
T^2=(4pi^2/Gm)(r^3)

 

[kuruman]

In problem 5 please explain your attempted strategy. It's not clear. Try using the given equation $g=\dfrac{Gm}{r^2}$ to find the ratio $\dfrac{g_{Jupiter}}{g_{Earth}}$.

In Problem 6a you use the appropriate equations but it looks like your calculation of the speed gives an incorrect number. Because you got that number wrong, 6b turned out wrong too.

Do you have a specific question to ask? It looks like you have gone over all this before and jotted down the correct answers in red.

 

[TextClick]

In hindsight, I don't understand my attempted strategy for Problem 5. Would you care to explain the process of solving it? Thanks!

This was a test, and the teacher did go over the answers, but I do not understand how to complete them.

 

[TextClick]

5) So I used your suggestion and plugged my values into the equation.

g=(6.67x10^-11)(1.79x10^27)/(7.04x10^7)=1.70x10^9 N

6) I redid 6a and got a different answer.

v=√(g.g7x10^-11)(5.97x10^24)/(6400100)=7888 m/s^2

Using that answer I redid 6b.

7888=2pi(6.4x10^6)/T=5096 s= 84.94 min

I am still confused by Problem 5, but for Problem 6, the answer is relatively close. It may have to do with rounding.

 

[PeroK]

5) So I used your suggestion and plugged my values into the equation.

g=(6.67x10^-11)(1.79x10^27)/(7.04x10^7)=1.70x10^9 N

There are two ways to do problem 5. You have tried the first way, which is to calculate the mass of Jupiter, calculate the radius of Jupiter and then calculate the surface gravity of Jupiter.

Your mistake is that you divided by the radius. It should be the square of the radius.

A quicker way was to relate the surface gravity of Jupiter to the surface gravity of Earth:

$g_J = \frac{GM_J}{R_J^2} = \frac{300}{11^2}\frac{GM_E}{R_E^2} = \frac{300}{121}g_E$

To do it this way, you don't need to know $G$ or the mass and size of either planet. You only need the Earth's surface gravity.

 

[PeroK]

v=√(g.g7x10^-11)(5.97x10^24)/(6400100)=7888 m/s^2

Using that answer I redid 6b.

7888=2pi(6.4x10^6)/T=5096 s= 84.94 min

but for Problem 6, the answer is relatively close. It may have to do with rounding.

It may have to do with putting the satellite only $100m$ and not $100km$ about the Earth's surface.