NEET Planet and Satellite MCQs & PYQs

Practice NEET Planet and Satellite Questions

Question 1:

moderate

A geostationary satellite is orbiting the earth at a height of $5R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is:

(2012 Pre)

For the geostationary satellite, $T_1 = 24\text{ hours}$, $r_1 = R + 5R = 6R$. For the second satellite, $r_2 = R + 2R = 3R$. Using Kepler's third law $T^2 \propto r^3$, we have $T_2 = T_1 \left(\frac{r_2}{r_1}\right)^{3/2} = 24 \left(\frac{3R}{6R}\right)^{3/2} = 24 \left(\frac{1}{2}\right)^{3/2} = 6\sqrt{2}\text{ hours}$.

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Question 2:

moderate

The mean radius of earth is R, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth’s surface is g. What will be the radius of the orbit of a geostationary satellite?

(1992)

Gravitational force provides the centripetal force: $$\frac{GMm}{r^2} = m\omega^2r$$.nSince $$g = \frac{GM}{R^2}$$, we can write $$GM = gR^2$$.nSubstituting this gives $$\frac{gR^2}{r^2} = \omega^2r \Rightarrow r^3 = \frac{gR^2}{\omega^2} \Rightarrow r = (\frac{R^2g}{\omega^2})^{\frac{1}{3}}$$.

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Question 3:

easy

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

(2005)

Kinetic energy $K = \frac{GMm}{2r}$ and Potential energy $U = -\frac{GMm}{r}$. The ratio of their magnitudes is $|K|/|U| = \frac{1}{2}$.

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Question 4:

moderate

A satellite of mass $m$ is orbiting the earth (of radius $R$) at a height $h$ from its surface. The total energy of the satellite in terms of $g_0$, the value of acceleration due to gravity at the earth’s surface, is:

(2016 – II)

Total energy of a satellite is the sum of kinetic and potential energy, given by $E = -\frac{GMm}{2(R+h)}$. Substituting $GM = g_0 R^2$, we obtain $E = -\frac{mg_0 R^2}{2(R+h)}$. Thus, option D is correct.

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Question 5:

easy

A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^6\text{ m}$ above the surface of earth. If earth’s radius is $6.38 \times 10^6\text{ m}$ and $g = 9.8\text{ m/s}^2$, then the orbital speed of the satellite is:

(2015 Re)

Orbital speed is calculated using $v = \sqrt{\frac{gR^2}{R+h}}$. Substituting the given values for $R$, $h$, and $g$, we get $v \approx 7.76\text{ km/s}$. Therefore, option B is correct.

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Question 6:

moderate

The radii of circular orbits of two satellites A and B of the earth, are $4R$ and $R$, respectively. If the speed of satellite A is $3V$, then the speed of satellite B will be:

(2010 Pre)

Orbital speed is inversely proportional to the square root of the radius ($v \propto \frac{1}{\sqrt{r}}$). Thus, $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}} = \sqrt{\frac{4R}{R}} = 2$, which means $v_B = 2 \times 3V = 6V$. Therefore, option C is correct.

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Question 7:

easy

A ball is dropped from a spacecraft revolving around the earth at a height of $120\text{ km}$. What will happen to the ball?

(1996)

When a ball is dropped from a spacecraft, it possesses the same orbital velocity as the spacecraft. Therefore, it continues to move with the same speed along the original orbit.

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