1. what is the term for the net force directed toward the center of an objects circular path?
a. circular force
b. centrifugal force
c. centripetal force
d. orbital force
2. if the distance between the earth and the moon is tripled, by what factor is the gravitational force between them changed?
a. 3
b. 1/3
c. 9
d. 1/9
3. if you swing a yo - yo in the circular path shown below, and release it at the dot, in what direction will it travel?
4. which of the vectors shown above represents the velocity of the ball?
a. a
b. b
c. c
d. d
5. which of the vectors shown above represents the acceleration of the ball?
a. a
b. b
c. c
d. d
6. two objects floating in space exert a gravitational force on one another. what happens to the magnitude of the gravitational force if one of their masses doubles and the distance between the two objects is kept the same?
a. the force doubles
b. the force quadruples
c. the force triples
d. the force remains the same
7. jupiters moon lo has a mass of 8.932 x 10²² kg and a radius of 1.822 x 10⁶ m. what is the gravitational field strength on the surface of lo?
a. 1.523 n/kg
b. 1.796 n/kg
c. 2.304 n/kg
d. 36.259 n/kg

Question

Accepted Answer

# Explanation:
## Step1: Recall centripetal force definition
Centripetal force is the net force directed toward the center of an object's circular path. Circular force is not a standard term. Centrifugal force is a fictitious force in a rotating frame. Orbital force is a general term for forces in orbits but not specific to the center - directed net force.
## Step2: Use Newton's law of gravitation $F = G\frac{m_1m_2}{r^{2}}$
Let the original distance be $r$, new distance $r'=3r$. Original force $F = G\frac{m_1m_2}{r^{2}}$, new force $F'=G\frac{m_1m_2}{(3r)^{2}}=G\frac{m_1m_2}{9r^{2}}=\frac{F}{9}$
## Step3: Understand object motion in circular path
When an object in circular motion is released, it travels in a straight - line tangent to the circular path at the point of release.
## Step4: Recall velocity direction in circular motion
The velocity of an object in circular motion is tangent to the circular path.
## Step5: Recall acceleration direction in circular motion
The centripetal acceleration of an object in circular motion is directed toward the center of the circular path.
## Step6: Use $F = G\frac{m_1m_2}{r^{2}}$
If $m_1$ becomes $2m_1$ and $r$ is constant. Original $F = G\frac{m_1m_2}{r^{2}}$, new $F'=G\frac{2m_1m_2}{r^{2}} = 2F$
## Step7: Use gravitational field strength formula $g=\frac{GM}{R^{2}}$
$G = 6.67\times10^{- 11}\space N\cdot m^{2}/kg^{2}$, $M = 8.932\times10^{22}\space kg$, $R = 1.822\times10^{6}\space m$
$$
\begin{align*}
g&=\frac{6.67\times 10^{-11}\times8.932\times 10^{22}}{(1.822\times 10^{6})^{2}}\
&=\frac{6.67\times8.932\times10^{11}}{3.32\times10^{12}}\
&=\frac{59.5\times10^{11}}{3.32\times10^{12}}\
&=\frac{59.5}{33.2}\
&\approx1.796\space N/kg
\end{align*}
$$

# Answer:
1. c. Centripetal force
2. d. $1/9$
3. Tangent to the circular path at the release point
4. c. C
5. a. A
6. a. The force doubles
7. b. $1.796\space N/kg$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall centripetal force definition

Step2: Use Newton's law of gravitation \(F = G\frac{m_1m_2}{r^{2}}\)

Step3: Understand object motion in circular path

Step4: Recall velocity direction in circular motion

Step5: Recall acceleration direction in circular motion

Step6: Use \(F = G\frac{m_1m_2}{r^{2}}\)

Step7: Use gravitational field strength formula \(g=\frac{GM}{R^{2}}\)

Answer: