The escape velocity is given by √2gR
Hence it's value Ve on the earth and Vm on the moon is
Ve = √2ge.Re
Vm = √2gm.Rm
Therefore , their ratio = Ve/Vm = √ge.Re/√gm.Rm = √6 x 10 = √60 = 8 nearly
First you need to caluclate the escape volocity. Calculating an escape velocity 1. Determine the mass and radius of the planet you are on. For Earth, assuming that you are at sea level, the radius is 6.38x10^6 meters and the mass is 5.97x10^24 kilograms. You will need the gravitational constant (G), which is 6.67x10^-11 N m^2 kg^-2. It is recommended to use metric units. 2. Using the above data, calculate the required velocity needed to exceed the planet's gravitational force. The object must have greater energy than the planet's velocity as follows: V(escape)= squareroot[(2GM)/r] where "M" is the mass of the earth, "G" is the gravitational constant(6.67x10^-11) and "r" is the radius from the center of the planet(6.38x10^6). 3. Accelerate the mass to the escape velocity. It is optimal to accelerate it perpendicular to the ground, assuming it is level. Accelerating the mass at an angle other than 90 degrees with respect to the ground will require a greater velocity such that V(escape)=V(actual)*sin(theta), where theta is the angle between the ground and the projected acceleration vector. 4. The escape velocity of Earth comes to about 11.2 kilometers per second or 25000 miles per hour from the surface.
They orbit around their common centre of gravity, the orbital radius and velocity of both (centripital force) is exactly enough to overcome the force of gravity between them.
the sun's radius is and half a million bigger than the radius of the sun.
The Earth's radius is only 4000 miles (approx.). The moon is approx. 250,000 miles away. The moon loses by a lot.
The Earth and the object exert a gravitational force on each other, but only the Earth's is big enough to measure. So, the formula for gravitational force include the distance from one body's surface to its center and the same for the other body. The length of the radius is directly proportional to the body's gravitational force.
For two bodies with equal radius, the more massive has the greater escape velocity. For two bodies with equal mass, the one with smaller radius has the greater escape velocity. Both conditions listed in the question indicate greaterescape velocity.
The Schwarzschild radius is a concept related to black holes. Given a body it is the radius such that, if all the mass of the body were squeezed (uniformly) within that sphere, then the escape velocity at the surface of the velocity would be equal to the speed of light.
no
m= mass of object v= velocity R= radius of the earth h= height from the surface to the escape velocity (mv2 ) ______ = G MEM R + h _____ (R+h)2 v2 =GME ______ (R +h)2 Ve = (√gR2 /(R+h)2 )
It is called the Schwarzschild radius
The escape velocity reaches the speed of light at the Schwarzshild radius.
The radius of Earth in kilometers is 6400 km.The earth is big in 6400 km in radius above its surface.
109 Actually, no. 109 would probably be for Jupiter. For Earth, hundreds of Earth's surface could fit in the sun's radius.
Force (newtons) = mass (kg) * acceleration (m/s/s) > Acceleration at earths surface radius = 9.82 m/s/s
MarsIts equatorial radius is 3,396.2 ± 0.1 km or 0.533 Earths,Its polar radius is 3,376.2 ± 0.1 km or 0.531 Earths,Its surface area is 144,798,500 km² or 0.284 Earths,Its volume is 1.6318 × 1011 km³ or 0.151 Earths,Its mass is 6.4185 × 1023 kg or 0.107 Earths.
At the radius of the earths surface, a body will gain roughly 10 metres per second velocity every second. (air resistance ignored) See google / acceleration due to gravity.
'Escape velocity' = sqrt(2 G m / r) =2.00114 x 10-8 km/sec = roughly 0.02 millimeter per second (rounded)