Detailed Answer

The definition needs to be learned. The formula is:

\[ F = G\frac{m_1 \cdot m_2}{r^2} \]

Detailed Answer

Field strength by definition is the force experienced by a unit mass. The force is:

\[ F = G\frac{m_1 \cdot m_2}{r^2} \]

When divided by \( m \), we get:

\[ g = G\frac{M}{r^2} \]

Detailed Answer

Field lines represent the direction a mass would accelerate if placed in that field.

Detailed Answer

It has both a direction (the field lines) and a magnitude, so it is a vector. It is important that when we add gravitational fields, we do so like they are vectors.

Detailed Answer

The way to derive this formula is really important to understand in this topic. It is based on equating the gravitational force with the centripetal force.

\[ m\omega^2 r = G\frac{m_1 \cdot m_2}{r^2} \]

\[ m\left(\frac{2\pi}{T}\right)^2 r = G\frac{m_1 \cdot m_2}{r^2} \]

\[ \frac{T^2}{r^3} = \frac{4\pi^2}{GM} \quad (= \text{constant}) \]

Detailed Answer

\( B \) is the first law.

Detailed Answer

The gravitational potential energy \(E_P\) of a two-body system is given by:

\[ E_P = - \frac{G m_1 m_2}{r} \]

Substituting the values:

\[ E_P = - \frac{(6.67 \times 10^{-11})(8.0)(12.0)}{3.0} = - 2.1 \times 10^{-9} \, \text{J}. \]

Detailed Answer

The gravitational field strength at the surface of a spherical body is given by:

\[ g = \frac{GM}{R^2}, \]

where \( M \) is the mass and \( R \) is the radius of the body.

For the planet:

• Mass of the planet: \( M_{\text{planet}} = 2M \),
• Radius of the planet: \( R_{\text{planet}} = 2R \) (since diameter is doubled).

Substituting into the formula for \( g_{\text{planet}} \):

\[ g_{\text{planet}} = \frac{G M_{\text{planet}}}{R_{\text{planet}}^2} = \frac{G (2M)}{(2R)^2}. \]

Simplify:

\[ g_{\text{planet}} = \frac{2G M}{4R^2} = \frac{1}{2} \cdot \frac{G M}{R^2}. \]

But \( \frac{G M}{R^2} = g \), the gravitational field strength at the surface of Earth.

Thus:

\[ g_{\text{planet}} = \frac{g}{2}. \]

Detailed Answer

The gravitational force between two masses is given by:

\[ F = \frac{G M_1 M_2}{r^2}, \]

where \( M_1 \) and \( M_2 \) are the masses, \( r \) is the distance between them, and \( G \) is the gravitational constant.

For sphere \( A \): \[ F_A = \frac{G M m}{r^2} = 60 \, \mathrm{N}. \]

For sphere \( B \):

\[ \text{The mass is } 2m, \text{ and the distance is } 2r. \] \[ F_B = \frac{G M (2m)}{(2r)^2}. \]

Simplify:

\[ F_B = \frac{2 G M m}{4r^2} = \frac{1}{2} \cdot \frac{G M m}{r^2}. \]

But \( \frac{G M m}{r^2} = F_A = 60 \, \mathrm{N} \). Therefore: \[ F_B = \frac{1}{2} \cdot 60 = 30 \, \mathrm{N}. \]