Question 24: Gravitation

Question

A planet has double the mass of the earth and double the radius. The gravitational potential at the surface of the Earth is $$\mathrm{V}$$ and the magnitude of the gravitational field strength is $$\mathrm{g}$$. The gravitational potential and gravitational field strength on the surface of the planet are

Potential
Field

A
V
$$\frac{g}{4}$$

B
2V
$$\frac{g}{2}$$

C
V
$$\frac{g}{2}$$

D
2V
$$\frac{g}{4}$$

Accepted Answer

To determine the energy required to move a body of mass $$m$$ from a circular orbit of radius $$3R$$ to a higher circular orbit of radius $$4R$$ around the Earth, we need to calculate the difference in total mechanical energy between the two orbits.
HereÃ¢ÂÂs how to do it step-by-step:
$$	extbf{Energy in a circular orbit:}$$
For an object in a circular orbit, the total mechanical energy (kinetic plus potential) is given by:
$$E = -\frac{1}{2}\frac{GMm}{r},$$
where
$$G$$ is the gravitational constant,
$$M$$ is the mass of the Earth, and
$$r$$ is the radius of the orbit.
$$	extbf{Energy at radius } 3R:$$
Substitute $$r = 3R$$ into the equation:
$$E_1 = -\frac{1}{2}\frac{GMm}{3R} = -\frac{GMm}{6R}.$$
$$	extbf{Energy at radius } 4R:$$
Substitute $$r = 4R$$:
$$E_2 = -\frac{1}{2}\frac{GMm}{4R} = -\frac{GMm}{8R}.$$
$$	extbf{Energy difference (required energy):}$$
The energy required to move the body is the difference between the energies of the two orbits:
$$\Delta E = E_2 - E_1.$$
Substitute our values:
$    \begin{aligned}    \Delta E &= \left(-\frac{GMm}{8R}ight) - \left(-\frac{GMm}{6R}ight) \    &= \frac{GMm}{6R} - \frac{GMm}{8R}.    \end{aligned}    $
To combine the fractions, find a common denominator:
$$\frac{GMm}{6R} = \frac{4GMm}{24R},\quad \frac{GMm}{8R} = \frac{3GMm}{24R}.$$
Therefore,
$$\Delta E = \frac{4GMm}{24R} - \frac{3GMm}{24R} = \frac{GMm}{24R}.$$
Thus, the energy required is
$$\frac{GMm}{24R}.$$
Comparing with the given options, the correct answer is:
Option B
$$\frac{GMm}{24R}.$$

Answer

C

Answer

D

Answer

A

Answer

B

Gravitation

Choose the correct answer: