## The minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R…

Q: The minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R is

Sol: From the law of conservation of energy

$\large -\frac{G M m}{R} + K.E = -\frac{G M m}{2 \times 3R}$

$\large K.E = \frac{G M m}{R}(1-\frac{1}{6})$

$\large = \frac{5}{6}(\frac{G M m}{R})$

## The mass of a spaceship is 1000 kg. It is to be launched from the earth surface out into free space. The value of ‘g’ and ‘R’ are…

Q: The mass of a spaceship is 1000 kg. It is to be launched from the earth surface out into free space. The value of ‘g’ and ‘R’ are 10 ms^(-2) and 6400 km respectively. The required energy for this work will be

Sol: to launch the spaceship out into free space, from energy conservation

$\large -\frac{G M m}{R} + E = 0$

$\large E = \frac{G M m}{R}$

$\large E = \frac{G M}{R^2} (m R) = m g R$

= 6.4 × 1010 J

## The gravitational potential difference between the surface of a planet and a point 20 m above it is 16 J/kg…

Q: The gravitational potential difference between the surface of a planet and a point 20 m above it is 16 J/kg. Calculate the work done in moving a 2 kg mass by 8 m on a slope of 60° from the horizontal.

Sol: The vertical height through which the body has to be raised = 8sin 60°=4√3 m.

The P.D for a distance of 20 m is 16 J/kg.

Hence, the P.D for a distance of 4√3 m for a mass of 2 kg $= \frac{4\sqrt{3}}{20}\times 16 \times 2$

= 11 J

## Two heavy spheres each of mass 100Kg and radius. 0.1m are placed 1m apart on a horizontal table. What is the gravitational field…

Q: Two heavy spheres each of mass 100Kg and radius. 0.1m are placed 1m apart on a horizontal table. What is the gravitational field and potential at the midpoint of the line joining their centres.

Sol: Gravitational field at the midpoint of the line joining their centres is given by

$\large E = \frac{GM}{(r/2)^2} – \frac{GM}{(r/2)^2}$

E = 0

Gravitational potential t the midpoint of the line joining their centres is given by

$\large V = -(\frac{GM}{(r/2)} + \frac{GM}{(r/2)}$

$\large V = -\frac{4GM}{r}$

$\large V = -\frac{4 \times 6.67 \times 10^{-11}\times 100}{1}$

=-2.7 × 10-8 j/kg.

## A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km/s. If 20% of its initial energy is lost…

Q: A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km/s. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ?
Mass of mars = 6.4 × 1023 kg; Radius of mars = 3395 km;

Sol: Sol: From the law of conservation of energy

$\large -\frac{G M m}{R} + (\frac{80}{100})\frac{1}{2} m v^2 = -\frac{G M m}{R+h} + 0$

$\large G M m(\frac{1}{R} – \frac{1}{R+h}) = 0.4 m (2 \times 10^3)^2$

R + h = 3,888.9 km

The required height up to which the rocket will go = 3,888.9 – 3,395 = 493.9 km.

## If a satellite is revolving around a planet of mass M in an elliptic orbit of semi major axis a, then show that the orbital speed…

Q: If a satellite is revolving around a planet of mass M in an elliptic orbit of semi major axis a, then show that the orbital speed of the satellite when it is at a distance r from the focus will be given by $v = \sqrt{GM (\frac{2}{r}-\frac{1}{a})}$

Sol: Total energy of the system is E = -GMm/2a

Which is conserved. So, KE + PE = -GMm/2a

At position ‘r’, orbital speed of the satellite is v .

Then, KE = (1/2)mv2 and PE = (-GMm)/r

$\large -\frac{G M m}{r} + \frac{1}{2} m v^2 = -\frac{G M m}{2a}$

$\large v = \sqrt{GM (\frac{2}{r}-\frac{1}{a})}$

## A particle is fired vertically upwards from the surface of earth reaches a height 6400km. Find the initial velocity of the particle.

Q: A particle is fired vertically upwards from the surface of earth reaches a height 6400km. Find the initial velocity of the particle.

Sol: TE. on the surface of the earth = TE. at the highest point

$\large -\frac{G M m}{R} + \frac{1}{2} m v^2 = -\frac{G M m}{R+h} + 0$

given, h = R = 6400 km

$\large v^2 = g h$

$\large v = \sqrt{g h}$

$\large v = \sqrt{10 \times 6400 \times 10^3}$

= 8 km/s