The definition needs to be learned. The formula is:
\[ F = G\frac{m_1 \cdot m_2}{r^2} \]
CloseField 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:
\[ g = G\frac{M}{r^2} \]
CloseField lines represent the direction a mass would accelerate if placed in that field.
CloseIt 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.
CloseThe 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.
\[ m\omega^2 r = G\frac{m_1 \cdot m_2}{r^2} \]
\[ m\left(\frac{2\pi}{T}\right)^2 r = G\frac{m_1 \cdot m_2}{r^2} \]
\[ \frac{T^2}{r^3} = \frac{4\pi^2}{GM} \quad (= \text{constant}) \]
CloseThe 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:
\[ E_P = - \frac{(6.67 \times 10^{-11})(8.0)(12.0)}{3.0} = - 2.1 \times 10^{-9} \, \text{J}. \]
CloseThe 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:
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:
\[ g_{\text{planet}} = \frac{g}{2}. \]
CloseThe 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: \[ F_B = \frac{1}{2} \cdot 60 = 30 \, \mathrm{N}. \]
CloseThe escape speed is given by:
\[ v_{\text{escape}} = \sqrt{\frac{2GM}{R}} \]
Substituting the given values:
\[ v_{\text{escape}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.0 \times 10^{24}}{7.0 \times 10^6}} \approx 9.7 \, \mathrm{km/s} \]
CloseCorrect answer: A
Gravitational potential energy (\( E_p \)) is given by:
\[ E_p = m \cdot V_g \],
where \( m \) is the mass of the satellite, and \( V_g \) is the gravitational potential.
The change in gravitational potential energy is:
\[ \Delta E_p = m \cdot \Delta V_g = m \cdot (V_{\text{final}} - V_{\text{initial}}) \]
Substitute the values:
\[ \Delta E_p = 2000 \, \text{kg} \cdot \left[ -25 \, \text{MJ} \ \text{kg}^{-1} - (-40 \, \text{MJ} \ \text{kg}^{-1}) \right] \],
\[ \Delta E_p = 2000 \, \text{kg} \cdot [15 \, \text{MJ} \ \text{kg}^{-1}] \].
Simplify:
\[ \Delta E_p = 30,000 \, \text{MJ} = 30 \, \text{GJ} \].
Since the gravitational potential increases (becomes less negative), the satellite moves away from Earth.
CloseCorrect answer: A
The weight of the satellite in orbit is given by:
\[ W_{\text{orbit}} = \frac{W}{r^2}, \]
where \( r \) is the ratio of the orbital radius to the Earth's radius.
Since the satellite is at \( r = 3R \), the weight becomes:
\[ W_{\text{orbit}} = \frac{W}{3^2} = \frac{W}{9} \approx 0.11 W. \]
The gravitational potential energy is given by:
\[ E_p = -\frac{GMm}{r}. \]
At \( r = 3R \), the potential energy becomes:
\[ E_{p, \text{orbit}} = \frac{E_p}{3} \approx 0.33 E_p. \]
Correct answer: A
The total energy (\( E_t \)) of a satellite in orbit is given by:
\[ E_t = -\frac{GMm}{2r}. \]
As the orbital distance (\( r \)) increases:
| Answer | Direction of movement of satellite | Change in gravitational potential energy / GJ |
|---|---|---|
| A | away from Earth | 30 |
| B | away from Earth | 10 |
| C | towards Earth | 10 |
| D | towards Earth | 30 |
| Answer | Gravitational Potential Energy | Weight |
|---|---|---|
| A | \( 0.33 E_p \) | 0.11 W |
| B | \( 0.5 E_p \) | 0.33 W |
| C | \( 0.25 E_p \) | 0.11 W |
| D | \( 0.33 E_p \) | 0.5 W |
| Answer | \( \Delta E_p \) | \( \Delta E_k \) | \( \Delta E_t \) |
|---|---|---|---|
| A | Increases | Decreases | Increases |
| B | Decreases | Increases | Decreases |
| C | Increases | Decreases | Decreases |
| D | Decreases | Decreases | Increases |