Physics · Science General

Escape Velocity and Rockets

92 Questions

Escape velocity and rockets questions cover spacecraft trajectories and gravitational influences. The topics include calculating maximum velocities and understanding multistage rocket mechanics. These physics concepts are essential for technical exams requiring scientific aptitude.

Escape velocityOrbital mechanicsRocket propulsionAtmospheric entry speedsGravitational pull

Escape Velocity and Rockets Questions

Multiple choice chemistry in daily life satellite launch vehicles rocket propellants

Suppose a rocket with an initial mass $M _0$ expels a mass $\delta m$ in the form of gases in time $\Delta t$, then the mass of the rocket after time t is : 

  1. $M _0$
  2. $M _0+\cfrac{\Delta m}{\Delta t}$
  3. $M _0-\cfrac{\Delta m}{\Delta t}$
  4. $M _0-\cfrac{\Delta m}{\Delta t}t$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The mass of the rocket decreases over time as fuel is expelled. If mass delta m is expelled in time delta t, the rate of mass loss is delta m / delta t. After time t, the mass lost is (delta m / delta t) * t.

Multiple choice lorentz transformation and muon experiment option a: relativity physics
A space traveler is moving at a speed of 0.6 times the speed of light as seen from the Earth's frame of reference. After one year passes on Earth, the space traveler will

  1. age more than one year

  2. age less than one year

  3. age exactly one year

  4. not age at all

  5. become younger
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given :   $v = 0.6 c$

The time elapsed on earth         $t  =1$ year.
Using relativity, time elapsed in the space traveler frame         $\tau = t \sqrt{1-\beta^2}$         where  $\beta = v/c  = 0.6$
$\therefore$   $\tau = 1 \times  \sqrt{1- (0.6)^2}  =1\times  0.8  = 0.8$ year                      $\implies$   $\tau < t$
Thus times elapsed in space traveler frame is less than that of earth, so space traveler will age less than one year.

Multiple choice lorentz transformation and muon experiment option a: relativity physics

An astronaut on a fast-moving spaceship appears to age only $1$ year to an outside observer, even though the person travels for $5$ years from the observer's perspective. The astronaut travels a distance of X during this time, from the observer's perspective.
Which of the following is true from the astronaut's perspective?

  1. The astronaut ages $5$ years during the trip

  2. The distance the astronaut travels is X

  3. The distance the astronaut travels is greater than X

  4. The distance the astronaut travels is less than X
  5. The astronaut ages more than $5$ years during the trip
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Time of journey from the person's perspective, $t = 5$ year
Thus distance traveled from the person's perspective, $X = vt =  5v$ 

Time of journey from the astronaut's perspective,  $t' = 1$ year
Thus distance traveled from the astronaut's perspective, $d' = vt' =  v$   $\implies$   $d'<X$ 

Multiple choice lorentz transformation and muon experiment option a: relativity physics

A person is watching a rocket with a astronaut inside move by at a speed near the speed of light. Which of the following statement is true?

  1. The length of the rocket is greater from the person's perspective than from the astronaut's perspective

  2. The length of the rocket is the same from the perspective of the person and the astronaut

  3. The length of the rocket is greater from the perspective of the astronaut than from the perspective of the person

  4. The person's length is greater, from his own perspective, as the rocket flies by, than it was before the rocket flew by

  5. The astronaut's length is greater, from his own perspective, as he flies by the person, than it was before he flew by the person
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Length of the moving rocket as seen by the person in rest frame       $L = L _o \sqrt{1-v^2/c^2}$     $\implies L<L _o$

where $L _o$ is the rest length of the rocket as seen by the astronaut in moving frame.
Thus length of the rocket is greater from the astronaut's perspective than from the person's perspective.

Multiple choice lorentz transformation and muon experiment option a: relativity physics

The length of the rod placed inside a rocket is measured as $1 m$ by an observer inside the rocket which is at rest. When the rocket moves with a speed of $36\times { 10 }^{ 6 }{ km }/{ hr }$ the length of the rod as measured by the same observer is :

  1. $0.997 m$
  2. $1.003 m$
  3. $1 m$
  4. $1.006 m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the rocket moves with some velocity, the observer inside it moves with the same speed. 

Hence the relative velocity of the rod with respect to the observer remains zero. 
Hence he observes the length to be the rest length $=1\ m$

Multiple choice relativistic mechanics option a: relativity physics

A person is watching a rocket with an astronaut inside move by at a speed near the speed of light.
Which of the following statements is true?

  1. The mass of the rocket is greater from the person's perspective than from the astronaut's perspective

  2. The mass of the rocket is the same from the perspective of the person and the astronaut

  3. The mass of the rocket is greater from the perspective of the astronaut than from the perspective of the person

  4. The person's mass is greater, from his own perspective, as the rocket flies by, than it was before the rocket flew by

  5. The astronaut's mass is greater, from his own perspective, as he flies by the person, than it was before he flew by the person
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass of the moving object as seen by the person in rest frame,  $m = \dfrac{m _o}{\sqrt{1-v^2/c^2}}$  $\implies m>m _o$

where  $m _o$ is the rest mass as seen by the astronaut in moving frame. 
Thus mass of rocket would be greater from the person's perspective than from the astronaut's perspective.

Multiple choice geography the earth and the graticule general idea about earth space around the earth is the earth round?

On what does the escape velocity of a moon depend?
I. Mass of celestial body.
II. The distance from the centre of mass to the escaping object.

  1. Only I

  2. Only II

  3. Both I and II

  4. Neither I nor II
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Escape velocity is the minimum speed needed for an object to escape from a massive body, in the sense of becoming neither on the surface nor in closed orbit of any radius no matter how great, without the aid of thrust, or suffering the resistance from friction. The escape velocity from Earth is about $11.186$ km/s$(6.951$ mi/s; $40,270$ km/h; $25,020$ mph$)$ at the surface.

Multiple choice social studies discovery of universe famous space scientists and astronauts space exploration space communications

How much speed is needed to escape the Earth's gravity?

  1. 9.2 Km/s

  2. 11.2 Km/s

  3. 12.4 Km/s

  4. 14.4 Km/s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In physics, escape velocity is the minimum speed needed for a free, non-propelled object to escape from the gravitational influence of a massive body. The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 36,700 ft/s; 25,020 mph; 21,744 km) at the surface.

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

If nitrogen gas molecule goes straight up with its rms speed at $0^o$C from the surface of the earth and there are no collisions with other molecules, then it will rise to an approximate height of:

  1. $18$ km
  2. $15$ km
  3. $12.38$ km
  4. $8$ km
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Molecular mass of Nitrogen molecule=$14$ g/mol

As nitrogen exists as ${N} _{2}$=28 g/mol=$0.028$ kg/mol
Also we know ${ v } _{ rms }=\sqrt { \dfrac { 3RT }{ M }  } $  where R= gas constant=8.31 bar/(K mol)=8.31$\times{10}^{5}$ Pa/(K mol)
T= temperature=${0}^{0}$ C=273 K
Also height $=\dfrac{{V}^{2} _{rms}}{2g}$

$=\dfrac { 3\times 8.31\times { 10 }^{ 5 }\times 273 }{ 2\times 9.81\times 0.028 } \ =12388\quad m=12.38\quad km$

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

A spaceship is launched into a circular orbit close to earth's surface. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is:
(Radius of earth $=6400km$ and $g=9.8m\quad { s }^{ -1 }$)

  1. $11.2km\quad { s }^{ -1 }$
  2. $8km\quad { s }^{ -1 }$
  3. $3.2km\quad { s }^{ -1 }$
  4. $1.5km\quad { s }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,
Radius of the earth, $R=6400km$
Acceleration due to gravity, $g=9.8m/s^2$
When a satellite is orbitting the earth its velocity is its orbital velocity given by orbital velocity: 
 ${ V } _{ 0 }=\sqrt { \dfrac { GM }{ R }  } =\sqrt { g } R$    $[\because g =\frac{GM}{R^2}]$
If it has to overcome gravitational pull then its velocity should be escape velocity escape velocity ${ V } _{ e }=\sqrt { \dfrac { 2GM }{ r }  } =\sqrt { 2gR } $
Additional velocity required is $V={ V } _{ e }-{ V } _{ 0 }=\sqrt { 2gR } -\sqrt { gR } =\sqrt { gR } \left( \sqrt { 2 } -1 \right) $
given that $R=64004m=64\times { 10 }^{ 6 }m$
$g=9.8m/{ s }^{ 2 }$
So, additional velocity will be:
$V=\sqrt { gR } \left( \sqrt { 2 } -1 \right) =\sqrt { 9.8\times 6.4\times { 10 }^{ 6 } } \left( \sqrt { 2 } -1 \right) $
$\Rightarrow V=3.28\times { 10 }^{ 3 }m/s$
Hence, the correct option is $(C)$
Multiple choice maths introduction to three dimensional geometry distance between two points in 3d distance between two points in space scalars and vectors

The equation of motion of a rocket are: $x=2t,y=-4t,z=4t,$ where the time $t$ is given in seconds and the coordinate of a moving point in kilometers. At what distance will the rocket be from the starting point $O(0,0,0)$ in $10$ seconds ?

  1. $60$ km
  2. $30$ km
  3. $45$ km

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Eliminating t from the given equation, we get the equation of the path $\dfrac{x}{2}=\dfrac{y}{-4}=\dfrac{z}{4}=t $

Thus the path of the Rocket represents a straight line passing through the origin for $t=10sec.$
we have $x=20,y=-40,z=40$
Let $\vec r=x\vec i+y\vec j+z\vec k$
$\Longrightarrow |\vec r|=\sqrt{{x^2}+{y^2}+{z^2}}=\sqrt{400+1600+1600}=60km$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A hydrogen balloon released on the moon would:

  1. climb up with an acceleration of $9.8 \ m/s^2$
  2. climb up with an acceleration of $9.8 \times 6 \ m/s^2$

  3. neither climb nor fall
  4. fall with an acceleration of $9.8/6 \ m/s^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As there is no atmosphere on the surface of the moon  so no bouyancy will act Hence it will fall with acceleration $9.8/6 m/s^2$

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

A stationary object is released from a point $P$ at a distance $3R$ from the centre of the moon which has radius $R$ and mass $M$. Which of the following gives the speed of the object on hitting the moon?

  1. $\left (\dfrac {2GM}{3R}\right )^{1/2}$
  2. $\left (\dfrac {4GM}{3R}\right )^{1/2}$
  3. $\left (\dfrac {GM}{3R}\right )^{1/2}$
  4. $\left (\dfrac {GM}{R}\right )^{1/2}$
Reveal answer Fill a bubble to check yourself
C Correct answer