A planet orbiting a distant star has radius 3.24 * 10^6 m. The escape speed for an object launched from this planet’s surface is 7.65 * 10^3 m/s. What is the acceleration due to gravity at the surface of the planet?

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Welcome back. Everyone in this problem, an asteroid in our solar system has a radius of about seven multiplied by 10 to the sixth meters and an escape velocity from its surface is 12 multiplied by 10 to the third meters per second. What would be its gravitational put A says it's 7 m per second squared, B eight C nine and D 10 and all are in units meters per second squared. Now, are we going to figure out the gravitational pull G given the information we already have that is the asteroid radius of seven multiplied by 10 to the sixth meters. So we can call that R and the escape velocity from its surface of 12 multiplied by 10 to the third meters per second. We can call that ve first. What do we know about gravitational pull? Well, recall that the acceleration due to gravity on any celestial body is given by the formula G equals GM divided by R squared where G represents the gravitational constant M represents the body's mass and R represents its radius. And we have all of the information except for its mass. So do we know anything else about the information in our problem statement that can help us to find the asteroids mass. Well, recognize we were also given the escape velocity and we know that the escape velocity is equal to the square root of two GM divided by R. So if we can manipulate our formula here to solve for M, then we should be able to substitute and solve for our gravitation output. So let's play around with our escape velocity formula for a bit. Now, if we square both sides and divide both sides by two, then we'll get V squared divided by two equal to GM divided by R. Now we could go ahead and continue to solve for our mass M. But notice that this expression is also in the formula for G. So we could substitute our value here half of the square of the escape velocity into our formula for that expression. And the alpha G because no, if we rewrite G as GM divided by R multiplied by one divided by R, now we can replace GM divided by R with V squared divided by two and multiply it by one divided by R, which means we're dividing V squared by R. OK. So the R formula ends up to be ve squared divided by two R. Now we have all that information so we can go ahead and solve our escape velocity is 12 multiplied by 10 to the third meters per second. So that's going to be squared And in our denominator, we're going to be multiplying two by seven, multiplied by 10 to the sixth meters the radius. Now, when we go ahead and solve here, then we should get G to be equal to 10.286 m per second squared or to one significant figure like the rest of our answers, approximately 10 m per second squared. Thus the gravitational pull of, of our asteroid would be answer choice. D thanks a lot for watching everyone. I hope this video helped.

A planet orbiting a distant star has radius 3.24 * 10^6 m. The escape speed for an object launched from this planet’s surface is 7.65 * 10^3 m/s. What is the acceleration due to gravity at the surface of the planet?

Verified Solution

Key Concepts

Escape Velocity

Gravitational Acceleration

Relationship Between Escape Velocity and Gravitational Acceleration