Question

Consider two planets of mass m and 2m,

Answer 1

Answer:

Part a)

$\frac{F_1}{F_2} = 10.125$

Part b)

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

$\frac{T_1}{T_2} = 9.54$

Explanation:

Part a)

As we know that the gravitational force is given as

$F = \frac{GMm}{r^2}$

so we will have to find the ratio of force on two planets due to star

so here we have

$\frac{F_1}{F_2} = \frac{m_1r_2^2}{m_2r_1^2}$

$\frac{F_1}{F_2} = \frac{m (4.5r)^2}{(2m) r}$

$\frac{F_1}{F_2} = 10.125$

Part b)

Orbital speed is given as

$v = \sqrt{\frac{GM}{r}}$

so the ratio of two orbital speed is given as

$\frac{v_1}{v_2} = \frac{r_2}{r_1}$

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

Time period is given as

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

so the ratio of two time period is given as

$\frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}}$

$\frac{T_1}{T_2} = \sqrt{\frac{4.5r^3}{r^3}}$

$\frac{T_1}{T_2} = 9.54$

Answer 2

Answer:

Explanation:

Mass of first planet, m1 = m

mass of second planet, m2 = 2m

distance of first planet, r1 = r

distance of second planet, r2 = 4.5 r

Let m be the mass of star.

(a) Force on first planet is given by

$F_{1}=\frac{GMm_{1}}{r_{1}^{2}}$

$F_{1}=\frac{GMm}{r^{2}}$     .... (1)

Force on second planet

$F_{2}=\frac{GMm_{2}}{r_{2}^{2}}$

$F_{2}=\frac{2GMm}{20.25r^{2}}$     .... (2)

Divide equation (1) by equation (2) ,we get

F1 : F2 = 10.125 : 1

(b) velocity for the first planet

$v_{1} = \sqrt{\frac{GM}{r_{1}}}$

$v_{1} = \sqrt{\frac{GM}{r}}$    .... (1)

velocity for the second planet

$v_{2} = \sqrt{\frac{GM}{r_{2}}}$

$v_{2} = \sqrt{\frac{GM}{4.5r}}$    .... (2)

Divide equation (1) by equation (2), we get

v1 : v2 = 2.12 : 1

(c) Orbital period for the first planet

$T_{1}=\frac{2\pi r_{1}}{v_{1}}$

$T_{1}=\frac{2\pi r}{v}$.... (1)

Orbital period for the second planet

$T_{2}=\frac{2\pi r_{2}}{v_{2}}$

$T_{2}=\frac{2\pi\times 4.5 r\times 2.12}{v}$.... (2)

Divide equation (1) by equation (2)

T1 : T2 = 1 : 9.55