Question 458: Gravitation

Question

Two spherical planets P and Q have the same uniform density r, masses M
P
and M
Q
and surface
areas A and 4A respectively. A spherical planet R also has uniform density r and its mass is (M
P
+ M
Q
). The escape velocities from the planets P, Q and R are V
P
, V
Q
and V
R
, respectively. Then

Accepted Answer

Escape velocity, expressed as
$$v_e = \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} = \sqrt{\frac{2 \mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^3 ho}{\mathrm{R}}}$$
can be simplified to
$$\sqrt{\frac{8 \pi \mathrm{G} ho}{3}} \mathrm{R}.$$
The surface area, $A_s$, is given by
$$A_s = 4 \pi R^2.$$
Considering planet Q's surface area is 4 times that of planet P:
$$\mathrm{A}_{\mathrm{Q}} = 4 \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{Q}}^2$$
$$\mathrm{~A}_{\mathrm{P}} = \mathrm{A} = 4 \pi \mathrm{R}_{\mathrm{P}}^2$$
therefore,
$$4 \mathrm{R}_{\mathrm{P}}^2 = \mathrm{R}_{\mathrm{Q}}^2.$$
Solving for $R_Q$ :
$$\Rightarrow \quad R_{\mathrm{Q}} = 2 \mathrm{R}_{\mathrm{P}} \quad 	ext{........(i)}.$$
For planet R, with mass $\mathrm{M}_{\mathrm{R}} = \mathrm{M}_{\mathrm{P}} + \mathrm{M}_{\mathrm{Q}}$:
$$ho \cdot \frac{4}{3} \pi \mathrm{R}_{\mathrm{R}}^3 = ho \cdot \frac{4}{3} \pi (\mathrm{R}_{\mathrm{P}}^3 + \mathrm{R}_{\mathrm{Q}}^3)$$
Since
$$\mathrm{R}_{\mathrm{Q}} = 2\mathrm{R}_{\mathrm{P}},$$
we have :
$$\mathrm{R}_{\mathrm{R}}^3 = \mathrm{R}_{\mathrm{P}}^3 + 8 \mathrm{R}_{\mathrm{P}}^3 = 9 \mathrm{R}_{\mathrm{P}}^3.$$
Thus,
$$\mathrm{R}_{\mathrm{R}} = (9)^{1/3} \mathrm{R}_{\mathrm{P}} \quad 	ext{........(ii)}.$$
From (i) and (ii), it follows that :
$$ R_R > R_Q > R_P.$$
Since
$$V_e \propto \mathrm{R},$$
it implies that :
$$\mathrm{V}_{\mathrm{R}} > \mathrm{V}_{\mathrm{Q}} > \mathrm{V}_{\mathrm{P}}.$$
Also,
$$\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{V}_{\mathrm{P}}} = \frac{\mathrm{R}_{\mathrm{R}}}{\mathrm{R}_{\mathrm{P}}} = 9^{1/3},$$
and
$$\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{Q}}} = \frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{Q}}} = \frac{1}{2}.$$

Answer

V
Q
> V
R
> V
P

Answer

V
R
> V
Q
> V
P

Answer

$${{{V_R}} \over {{V_P}}} = 3$$

Answer

$${{{V_P}} \over {{V_Q}}} = {1 \over 2}$$

Gravitation

Two spherical planets P and Q have the same uniform density r, masses M P and M Q and surface areas A and 4A respectively. A spherical planet R also has uniform density r and its mass is (M P + M Q ). The escape velocities from the planets P, Q and R are V P , V Q and V R , respectively. Then

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