Question

1. a cow and an elephant each of mass 2000 kg attract each other gravitationally with a force of 1×10⁻⁵n. how far apart are they? (a) 1.145 m (b) 3.785 m (c) 5.156 m (d) 10 m (e) none of the above 24. the force of gravity between two planets depends on their (a) masses and distance apart (b) planetary atmospheres (c) rotational motions (d) all of the above

Explanation:

Step1: Recall gravitational - force formula

The gravitational - force formula is $F = G\frac{m_1m_2}{r^2}$, where $F$ is the gravitational force, $G=6.67\times 10^{- 11}\ Nm^2/kg^2$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between them. Given $m_1 = m_2=2000\ kg$ and $F = 1\times 10^{-5}\ N$.

Step2: Rearrange the formula for $r$

From $F = G\frac{m_1m_2}{r^2}$, we can solve for $r$:
\[

$$\begin{align*} r^2&=G\frac{m_1m_2}{F}\\ r&=\sqrt{G\frac{m_1m_2}{F}} \end{align*}$$

\]

Step3: Substitute the values

Substitute $G = 6.67\times 10^{-11}\ Nm^2/kg^2$, $m_1 = m_2 = 2000\ kg$, and $F = 1\times 10^{-5}\ N$ into the formula:
\[

$$\begin{align*} r&=\sqrt{\frac{6.67\times 10^{-11}\times2000\times2000}{1\times 10^{-5}}}\\ &=\sqrt{\frac{6.67\times10^{-11}\times4\times10^{6}}{1\times 10^{-5}}}\\ &=\sqrt{6.67\times4\times10^{-11 + 6+5}}\\ &=\sqrt{26.68}\\ &\approx5.16\ m \end{align*}$$

\]
For question 24:
The gravitational force between two objects (including planets) is given by Newton's law of universal gravitation $F = G\frac{m_1m_2}{r^2}$, which depends on their masses ($m_1$ and $m_2$) and the distance ($r$) between them. Planetary atmospheres and rotational motions do not affect the gravitational - force formula.

Answer:

1. c. 5.156 m
2. a. masses and distance apart